I am flying from Chicago to Istanbul and my flight will layover in Rome for an hour and a half. I am holding a US green card and am traveling with Iraqi passport. I am not sure if I will require an Italian transit visa or will I be fine since am not leaving the transit area? I tried the Italy embassy support line and they keep telling me to check the website and there was not a clear answer for that.
I found many related questions on this site but I am still not sure about the rules. How can I decide if I need a visa to transit? Schengen members as of May 2021 are as follows:  Austria Belgium Czech Republic Denmark (excluding Greenland and the Faroe Islands - but an open border with the Schengen Area is maintained) Estonia Finland France (excluding overseas departments and collectivities) Germany Greece Hungary Iceland Italy Latvia Liechtenstein Lithuania Luxembourg Malta Netherlands (excluding Aruba, Curaçao, Sint Maarten and the Caribbean Netherlands) Norway (excluding Svalbard) Poland Portugal Slovakia Slovenia Spain (except Ceuta and Melilla) Sweden Switzerland
Why "hide" Luke from his father somewhere Anakin knows? I want to subtract the current frame from a previous frame, I have a image sequence and I want blender to subtract the current frame from the frame with an offset of -4. If I am on frame 5 so blender will subtract and use frame 1. I've been battling the following two questions for more than a day today.  Write a regular expression (comprised of {a, b}) that contain at least two b and do not contain abb. Write a regular expression (comprised of {a, b}) that contain abba and do not contain baa.  For the first part I wrote a basic one to start off with that doesn't handle the must two-b: b* OR a* OR (a+ AND b)* OR a*  I ran a few &quot;truth tables&quot; inputs over it and it seems ok. But the moment I add the requirement, things go haywire. Mostly because when pushing in those &quot;b&quot;s, you can suddenly use the OR to break things up. Same thing on the second one.... Any help would be greatly appreciated! Norah Why do we use the object instead of the subject pronoun in constructions like "stupid me"? I'm trying to find out how come we say lucky me and stupid us rather than lucky I and stupid we. My understanding is that this is not a recent invention, but a relic from the distant past where it was a dative construction. However, I don't see the reason for using any case other than the nominative in this construction to begin with.        He went surfing. → Lucky him went surfing.   I got pulled over. → Stupid me got pulled over.   They bought a house. → Thoughtless them bought a house.      This is rather intriguing. After thinking about it for a while, I couldn't come up with any other language, Germanic or otherwise, where simply introducing an adjective would dictate a change of grammatical case. This appears to be an English peculiarity. How far can it be traced back? Can anyone comment on its background? Failed to fetch jessie backports repository I'm using a docker image as a base for my own development that adds the jessie backports repository in its Dockerfile and uses that to install a dependency. This image uses the following command to add the repository:  echo "deb http://ftp.debian.org/debian jessie-backports main" &gt;&gt; /etc/apt/sources.list   The problem is that fetching packages from the backports repository now  fails with the following error (this used to work previously):  W: Failed to fetch http://ftp.debian.org/debian/dists/jessie-backports/main/binary-amd64/Packages 404  Not Found  W: Failed to fetch http://deb.debian.org/debian/dists/jessie-updates/main/binary-amd64/Packages  404  Not Found   I looked on that server, and those paths are indeed not present there.  I tried to figure out on the Debian backports site whether this particular repository should still be available, and I didn't find any indication that this was deprecated or something like that.  Is this a temporary issue with the repository, or is the jessie-backports repository not available anymore? And if this is not a temporary issue, what options do I have to use this or an equivalent repository without upgrading to the newer Debian stable version? How to remove version and thank you message from the admin footer I've used following codes and no luck  1  //Hide admin footer from admin function change_footer_admin () {     return ' '; } add_filter('admin_footer_text', 'change_footer_admin', 9999);  function change_footer_version() {     return ' '; } add_filter( 'update_footer', 'change_footer_version', 9999);   2  function wpbeginner_remove_version() { return ''; } add_filter('the_generator', 'wpbeginner_remove_version');   3  function my_footer_shh() {     remove_filter( 'update_footer', 'core_update_footer' );  } add_action( 'admin_menu', 'my_footer_shh' );   Still I see those texts and version on my footer.   So what is the accurate way to remove these? I've added these code snippets into my child theme function.php Search which man sections contain a given entry I've run across (personally or by proxy) a bunch of storage media that exhibited physical problems during their life. Hard disks, solid state drives and other flash storage media all fail after some time. The situation is worse with USB flash drives and flash cards because of the large amount of suppliers with inferior flash cell and flash controller quality.  How can I reliably detect if a storage drive suffers from physical damage?  Note that this question isn't about file system integrity (which can be checked with ). Express symmetric polynomial $\prod_{i < j} (X_i+X_j)$ in terms of elementary symmetric functions Exercise: Define a polynomial $\Sigma(X_1,\ldots,X_n)$ as \begin{align*} \Sigma(X_1,\ldots,X_n) = \prod_{i &lt; j} (X_i+X_j) \end{align*} This is a symmetric polynomial, quite clearly. I want to express $\Sigma$ as a polynomial in the elementary symmetric functions, for $n=2,3,4$. Case $n=2$ is trivial: $\Sigma(X_1,X_2)= \sigma_1 = X_1+X_2$. Case $n=3$ is easy. I write $\Sigma$ as follows: \begin{align*} \Sigma(X_1,X_2,X_3) = (\sigma_1 - X_1)(\sigma_1 - X_2)(\sigma_1 - X_3), \end{align*} then apply  formally, obtaining \begin{align*} \Sigma(X_1,X_2,X_3) = \sigma_1^3 - \sigma_1 \sigma_1^2 + \sigma_2 \sigma_1 - \sigma_3 = \sigma_2 \sigma_1 - \sigma_3. \end{align*} The case $n=4$ seems harder to me, and I'm unable to work it out. I hope there is some trick to avoid unnecessary computations. My attempt is to write down $\Sigma$ completely: \begin{align*} \Sigma = (X_1+X_2)(X_3+X_4)(X_1+X_4)(X_2+X_3)(X_2+X_4)(X_1+X_3) \end{align*} and then multiply factors in groups of two, obtaining: \begin{align*} \Sigma = (\sigma_2 - X_1X_2-X_3X_4)(\sigma_2 - X_1X_4 - X_2X_3)(\sigma_2 - X_2X_4-X_1X_3), \end{align*} but it doesn't seem to help much. How can I solve the exercise? How do I extract all the external links of a web page and save them to a file? With a two-party system, our nation will divide the people (who/whom)   most need to be brought together.   Do I use who or whom for this sentence?  I think that "people" is the direct object and warrants the use of "whom", but I want to make sure I'm right. Automatically detect edit abuse and exclude from badges/flag for mod review has 5 pages of 1 character edits (including an edit to one of my posts), apparently in a massive attempt to earn the Copy Editor badge.  Should Copy Editor, etc., ignore 1 character whitespace-only edits?    If we still want to count minor edits (which may be useful), could we at least detect edit abuse via some secret algorithm? ~20+ one character whitespace-only edits in the course of an hour should at least raise a flag of some sort.    If whitespace character are too hard to count, can we at least assume that 20 non-tag edits in 20 minutes (less than a minute per edit) is a good signal of abuse? The specific algorithm should be secret, obviously, but I think these sorts of edit abuses shouldn't be too difficult to detect. caching JavaScript files Which is the best method to make the browser use cached versions of js files (from the serverside)? I read about Java's type erasure .  When does type erasure occur? At compile time or runtime? When the class is loaded? When the class is instantiated?  A lot of sites (including the official tutorial mentioned above) say type erasure occurs at compile time. If the type information is completely removed at compile time, how does the JDK check type compatibility when a method using generics is invoked with no type information or wrong type information?  Consider the following example: Say class A has a method, empty(Box&lt;?&nbsp;extends&nbsp;Number&gt;&nbsp;b). We compile A.java and get the class file A.class.  public class A {     public static void empty(Box&lt;? extends Number&gt; b) {} }     public class Box&lt;T&gt; {}   Now we create another class B which invokes the method empty with a non-parameterized argument (raw type): empty(new Box()). If we compile B.java with A.class in the classpath, javac is smart enough to raise a warning. So A.class has some type information stored in it.  public class B {     public static void invoke() {         // java: unchecked method invocation:         //  method empty in class A is applied to given types         //  required: Box&lt;? extends java.lang.Number&gt;         //  found:    Box         // java: unchecked conversion         //  required: Box&lt;? extends java.lang.Number&gt;         //  found:    Box         A.empty(new Box());     } }   My guess would be that type erasure occurs when the class is loaded, but it is just a guess. So when does it happen? Ok, so I have an index.php file which has to process many different file types.  how do I guess the filetype based on the REQUEST_URI.  If I request http://site/image.jpg, and all requests redirect through index.php, which looks like this  &lt;?php    include('/www/site'.$_SERVER['REQUEST_URI']); ?&gt;   How would I make that work correctly?  Should I test based on the extension of the file requested, or is there a way to get the filetype? Let $F$ be a field. $(x-c)|(p(x)-p(c))$.   So if $x-c|p(x)=q(x)-\frac{r}{x-c}$ so for $x-c|p(x)-p(c)=s(x-c)-\frac{r}{x-c}$. Is that correct? Or is there a different way to go about proving this? I have proved that $x-c|p(x)$ has a remainder of $p(c)$ so It wuld make sense that $p(x)-p(c)$ would be $x-c$. "The train will leave" vs. "is going to leave" vs. "leaves" vs. "is leaving" Is it ethical to withdraw a paper after acceptance in order to resubmit to a better journal? Notepad doesn't open files created by Gedit? Community Wiki posts don't change usernames if the user does