Do I require transit visa to stay at Narita International Airport?
Layover at Narita, Japan. Do I need a transit visa?
Flagging history doesn't show anything when the only flags being cast are spam flags Today I've helped flag a lot of spam for this: .  When I've visited my profile on WebApps to see my flag history of flag, the information about my flagging history was missing.  Note that these spam flags were my only flags on this site.    Normally, I should have more information like when I'm looking at my Stack Overflow account.    Is this a bug? Easy curves in TikZ I find myself needing to draw lots of elegantly curved paths in TikZ. Ideally, I'd just specify a series of points, and TikZ would calculate the extra data itself to draw a nice series of curves passing smoothly through these points, perhaps with an optional "looseness" parameter that I could specify. But the only way I can find to draw nice curves is by explicitly giving control points, or by manually specifying in and out angles.  I can think up a simple algorithm to do this, which would certainly be within TikZ's power to perform: just choose the in and out angles in a simple fashion based on the relative angles between each adjacent pair of line segments.  Is something like this already built-in? Or can someone cook something up that does the job?  Edit: Jake has provided an answer using the plot [smooth] functionality. This is almost perfect! But it can't do what I need, because it doesn't let me specify tangent angles manually where needed, which is especially important at the beginning and end of the curve. I would have thought this would be a natural and straightforward addition to the existing plot [smooth] algorithm: for every coordinate, an optional angle should be able to be specified as an argument, which if supplied is treated as the tangent angle for the curve at that point. And while we're at it, it wouldn't hurt to also allow the tension to be modified along the path.  A minimal extension to the algorithm would just accept two optional parameters, for the curve tangent at the beginning and end. I am trying to analyze data based on a sample size of around 50 and would appreciate advice on which statistical test would be most appropriate. Participants, on a single questionnaire, were asked to complete two different sections. Variable 1: rating Likert items on a scale from 1-5 (strongly agree to disagree). Variable 2: rating Likert items of a validated tool on a scale of 1-9, which is averaged to obtain a single number.  My research question is: Do those who 'agree' with the statement in the first section significantly differ in terms of their scores on the variable 2, compared to those who 'disagree'? (ex. Do participants tend to 'agree' with "I like to read books for pleasure." have higher IQs (variable 2) than those who 'disagree'?)  I was advised by a colleague to run a one-way ANOVA with variable 1 (categorical) as IV and variable 2 (continuous) as DV.   Question 1: My independent variable is the three groups: 'agree', 'neutral', and 'disagree.' (I combined strongly agree with agree because of low frequencies and likewise for strongly disagree and disagree.) However, there are many more participants in the 'agree' category than in the 'disagree' category. The distribution is, on average, around 44 agrees, 3 neutrals, and 3 disagrees. Do the (very) unequal sample sizes prevent me from performing ANOVA, or any other tests? Which test would be most appropriate for this analysis?  Question 2: I would like to run analyses on all of the Likert items, which is more than 10 statements. (ex. Seeing how IQs correlate with agreement to many statements including “I like to read books for pleasure,” “I watch a lot of TV,” “I exercise more than 30 minutes each week,” etc.) I was planning to do a Bonferroni correction to adjust for the multiple tests. Would this be correct, and help make my analyses more robust? Note: The continuous data are not normally distributed (there is negative skewness), but I seem to be able to correct the skewness using logarithmic transformation.  Any advice or suggestions for papers to read would be appreciated. Thank you! Is there a way for non-root processes to bind to "privileged" ports on Linux? It's very annoying to have this limitation on my development box, when there won't ever be any users other than me.  I'm aware of , but none of them do exactly what I want:    (The version in Debian testing, 1.0, only supports IPv4)  (the "nat" table is not yet implemented for ip6tables, the IPv6 version of iptables) sudo (Running as root is what I'm trying to avoid) SELinux (or similar). (This is just my dev box, I don't want to introduce a lot of extra complexity.)   Is there some simple sysctl variable to allow non-root processes to bind to "privileged" ports (ports less than 1024) on Linux, or am I just out of luck?  EDIT: In some cases, you can  to do this. How do I change my Minecraft username? Text gets stretched or disappears when scrolling When scrolling in the Stack Exchange Android app, the text gets stretched vertically, code disappears, and sometimes the entire post content disappears.  This doesn't make the app completely useless, but it's pretty close.   Motorola Razr MAXX, Android 4.0.4. Python string.join(list) on object array rather than string array In Python, I can do:  &gt;&gt;&gt; list = ['a', 'b', 'c'] &gt;&gt;&gt; ', '.join(list) 'a, b, c'   Is there any easy way to do the same when I have a list of objects?  &gt;&gt;&gt; class Obj: ...     def __str__(self): ...         return 'name' ... &gt;&gt;&gt; list = [Obj(), Obj(), Obj()] &gt;&gt;&gt; ', '.join(list) Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in &lt;module&gt; TypeError: sequence item 0: expected string, instance found   Or do I have to resort to a for loop? How to render from Eevee in Blender 2.8? I have the Blender 2.8 version from the 21st Sep 2018 I was looking forward to render a test sequence but cannot find a render button anywhere? Does anyone know where to find it? How do I quickly switch between weapons and spells? When  I see him quickly change his weapons and spells during combat. When he later goes inside his items, you can see a second smaller arrow next to the weapon that you can quickly change to in combat. How can I set such an alternative so that I can quickly switch weapon?  The screen doesn't seem to intuitively explain what key I need to use to do that... biblatex: displaying all authors of multi-author works in the bibliography I'm using the  package to add citations and a bibliography to my LaTeX document. I've noticed that only the first author plus "et al." is displayed for works with more than three authors. That's fine with me for in-text-citations, but I'd rather have the complete author information in the bibliography. How can I do that? Determine all of the accumulation points of the following sets in $\mathbb{R}^1$ and decide whether the sets are open or closed or neither.  The set $X$ of all numbers of the form $2^{-n} + 5^{-m}$ where $m,n = 1,2,\ldots$.   Claim: The set of accumulation points is the following: $$\left\{\frac{1}{2^n}: n \in \mathbb{N} \right\} \cup \left\{\frac{1}{5^n}: n \in \mathbb{N}\right\} \cup \{0\}.$$ Let $n \in \Bbb{N}$ and $r&gt;0$, and consider the open neighbourhood $B\left(\frac{1}{2^n},r\right) = \left(\frac{1}{2^n} - r,\frac{1}{2^n} + r\right)$. For the given $r$ we can find $m$ large enough such that $5^{-m} &lt; r \implies \frac{1}{2^n} + \frac{1}{5^m} &lt; \frac{1}{2^n} + r$, and $\frac{1}{2^n} - \frac{1}{5^m} &gt; \frac{1}{2^n} - r$; therefore $5^{-m} + 2^{-n} \in B\left(\frac{1}{2^n},r\right)$. A similar argument shows that each element of $\{5^{-m}: m \in \mathbb{N}\} \cup \{0\}$ is an accumulation point of $X$. Now consider $y \in X$. $B(y,r)$ isn't a subset of $X$ since for all $r &gt; 0$, $B(y,r) \subset (\mathbb{R} - \mathbb{Q}) \not\subset \Bbb{Q}$, hence $B(y,r) \not\subset X$. Tikzcd conflict with babel and cd I have this problem: when I try to compile a diagram I get the following message:     I think the culprit is a tikzcd arrow in ell 1-1   But I have not problem when, in the tikzlibrary, I just use cd, matrix and backgrounds. The problem is when using babel and cd at the same time.  \documentclass{article} \usepackage{amsfonts,amssymb,amsmath,tikz} \usepackage[spanish]{babel} \usepackage[latin1]{inputenc} \usepackage[mathcal]{euscript}  \usetikzlibrary{babel,cd,matrix,backgrounds}  \begin{document} \begin{center}   \begin{tikzcd}     X\rar[mapsto]{T}&amp;Y   \end{tikzcd} \end{center}  \end{document} Do ground seeds contain oil? Convert XML/HTML Entities into Unicode String in Python I'm trying to connect to a sandbox using the sfdx cli and the JWT flow through force:auth:jwt:grant. According to , we can use the same connected app information from Production to connect to our Scratch Org.  Is it something that should work for sandbox too ? I'm getting a "Failed: Not approved" error.  Connected app in sandbox comes from the refresh but has a different consumer key. The Permission set is set correctly on my user.  Is there a way to connect to a recently refreshed sandbox using the JWT flow (without manually changing something like the consumer key) ? Share your experiences on exploring iOS 7 with us..  Whether you found hidden or little-known feature of iOS 7 or can explain how a big feature makes a difference, we're looking for objective reports of how the new OS affects your workflow.  Also provide details on how to achieve that feature, and if possible, include a relevant image too (hold Home and Lock for a screenshot on your iOS device).    Please post one tip per answer. Please also check to see if your answer has already been posted - duplicate answers will be deleted in favour of the original answer on any given topic. To search answers for this question use inquestion:101382 (or inquestion:this, directly from the question page) in addition to your search terms in the search box in the upper right hand corner of this page. Graph Proof by induction. Can you prove via induction that there exists a node in a directed graph of n nodes that can be reached in at most two edges from every other node in the graph. Every node in the graph is required to have an edge with every other node that either points out or in (For every pair of nodes X and Y, either X points to Y or Y points to X). Confirming that my own question is a duplicate by voting to close should have immediate effect How can I prove: if $p $ is prime and $n>1$, then $ p^{\frac1n} $ is irrational? Why multiple capacitors in parallel? Wondering why you would choose to use smaller capacitors in parallel vs one bigger one.  Here's an example:      Is this a component size and track width choice?  Thanks!