description: You have a string s and a chip, which you can place onto any character of this string. 
  
  After placing the chip, you move it to the right several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is i, you move it to the position i + 1. Of course, moving the chip to the right is impossible if it is already in the last position.
  
  After moving the chip to the right, you move it to the left several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is i, you move it to the position i - 1. Of course, moving the chip to the left is impossible if it is already in the first position.
  
  When you place a chip or move it, you write down the character where the chip ends up after your action. For example, if s is abcdef, you place the chip onto the 3-rd character, move it to the right 2 times and then move it to the left 3 times, you write down the string cdedcb.
  
  You are given two strings s and t. Your task is to determine whether it's possible to perform the described operations with s so that you write down the string t as a result.
  
  Input
  
  The first line contains one integer q (1 ≤ q ≤ 500) — the number of test cases.
  
  Each test case consists of two lines. The first line contains the string s (1 ≤ |s| ≤ 500), the second line contains the string t (1 ≤ |t| ≤ 2 ⋅ |s| - 1). Both strings consist of lowercase English characters.
  
  It is guaranteed that the sum of |s| over all test cases does not exceed 500.
  
  Output
  
  For each test case, print "YES" if you can obtain the string t by performing the process mentioned in the statement with the string s, or "NO" if you cannot.
  
  You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
  
  Example
  
  Input
  
  
  6
  abcdef
  cdedcb
  aaa
  aaaaa
  aab
  baaa
  ab
  b
  abcdef
  abcdef
  ba
  baa
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  YES
  NO
  
  Note
  
  Consider the examples.
  
  The first test case is described in the statement.
  
  In the second test case, you can place the chip on the 1-st position, move it twice to the right, and then move it twice to the left.
  
  In the fourth test case, you can place the chip on the 2-nd position, and then don't move it at all.
  
  In the fifth test case, you can place the chip on the 1-st position, move it 5 times to the right, and then finish the process.
solution: import java.io.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
  util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
  import static java.lang.System.*;
  public class Main
  {
      char a[];
      char b[];
      int aa;
      int bb;
      public void tq()throws Exception
      {
          String y="YES";
          String n="NO";
          st=new StringTokenizer(bq.readLine());
          int tq=i();
          sb=new StringBuilder(2000000);
          o:
          while(tq-->0)
          {
              a=s().toCharArray();
              b=s().toCharArray();
              aa=a.length;
              bb=b.length;
              for(int x=0;x<aa;x++)
              {
                  if(a[x]==b[0])
                  {
                      if(r(x,0)||l(x,0))
                      {
                          sl(y);
                          continue o;
                      }
                  }
              }
              sl(n);
          }
          p(sb);
      }
      boolean r(int i,int j)
      {
          if(j==bb)return true;
          if(i==aa)return false;
          if(a[i]==b[j])return r(i+1,j+1)||l(i-1,j+1);
          return false;
      }
      boolean l(int i,int j)
      {
          if(j==bb)return true;
          if(i==-1)return false;
          if(a[i]==b[j])return l(i-1,j+1);
          return false;
      }
      int di[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
      long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
          MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
          StringBuilder sb;public static void main(String[] a)throws Exception{new Main().tq();}int[] so(int ar[])
      {Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;
          x++)ar[x]=r[x];return ar;}long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)
              r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}char[] so(char ar[]) {Character
          r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)
              ar[x]=r[x];return ar;}void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.
          append(s);}void s(char s){sb.append(s);}void s(double s){sb.append(s);}void ss(){sb.append(' ');}void sl
      (String s){sb.append(s);sb.append("\n");}void sl(int s){sb.append(s);sb.append("\n");}void sl(long s){sb
          .append(s);sb.append("\n");}void sl(char s) {sb.append(s);sb.append("\n");}void sl(double s){sb.append(s)
      ;sb.append("\n");}void sl(){sb.append("\n");}int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
      long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
      {return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
      {while(b>0){int c=a%b;a=b;b=c;}return a;}boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=
      s.charAt(j--))return false;return true;}boolean[] si(int n) {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]
      =true;for(int x=4;x<=n;x+=2)bo[x]=true;for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;
          y+=vv)bo[y]=true;}}return bo;}long mul(long a,long b,long m) {long r=1l;a%=m;while(b>0){if((b&1)==1)
              r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}int i()throws IOException{if(!st.hasMoreTokens())st=new
          StringTokenizer(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
      {if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
      s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
      double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
          parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
      {out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
      p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
      {out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
          println(p);}void pl(boolean p){out.println(p);}void pl(){out.println();}void s(int a[]){for(int e:a)
          {sb.append(e);sb.append(' ');}sb.append("\n");}void s(long a[]){for(long e:a){sb.append(e);sb.append(' ')
      ;}sb.append("\n");}void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append
          ("\n");}}void s(char a[]){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}void s(char ar[][])
      {for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}int[] ari(int n)throws
          IOException {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());for(int x=0;
                  x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}int[][] ari(int n,int m)throws
          IOException {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if (!st.hasMoreTokens())st=new StringTokenizer
          (bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}long[] arl
      (int n)throws IOException {long ar[]=new long[n];if(!st.hasMoreTokens()) st=new StringTokenizer(bq.readLine())
      ;for(int x=0;x<n;x++)ar[x]=Long.parseLong(st.nextToken());return ar;}long[][] arl(int n,int m)throws
          IOException {long ar[][]=new long[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens()) st=new
          StringTokenizer(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
      String[] ars(int n)throws IOException {String ar[] =new String[n];if(!st.hasMoreTokens())st=new
              StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken();return ar;}double[] ard
          (int n)throws IOException {double ar[] =new double[n];if(!st.hasMoreTokens())st=new StringTokenizer
          (bq.readLine());for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}double[][] ard
      (int n,int m)throws IOException{double ar[][]=new double[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens())
          st=new StringTokenizer(bq.readLine());for(int y=0;y<m;y++) ar[x][y]=Double.parseDouble(st.nextToken());}
      return ar;}char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new
          StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}char[][]
      arc(int n,int m)throws IOException {char ar[][]=new char[n][m];for(int x=0;x<n;x++){String s=bq.readLine();
          for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}void p(int ar[]){StringBuilder sb=new StringBuilder
          (2*ar.length);for(int a:ar){sb.append(a);sb.append(' ');}out.println(sb);}void p(int ar[][])
      {StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(int a[]:ar){for(int aa:a){sb.append(aa);
          sb.append(' ');}sb.append("\n");}p(sb);}void p(long ar[]){StringBuilder sb=new StringBuilder
          (2*ar.length);for(long a:ar){ sb.append(a);sb.append(' ');}out.println(sb);}void p(long ar[][])
      {StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(long a[]:ar){for(long aa:a){sb.append(aa);
          sb.append(' ');}sb.append("\n");}p(sb);}void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;
              StringBuilder sb=new StringBuilder(c);for(String a:ar){sb.append(a);sb.append(' ');}out.println(sb);}
      void p(double ar[]){StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
          sb.append(' ');}out.println(sb);}void p(double ar[][]){StringBuilder sb=new StringBuilder(2*
          ar.length*ar[0].length);for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n")
      ;}p(sb);}void p(char ar[]){StringBuilder sb=new StringBuilder(2*ar.length);for(char aa:ar){sb.append(aa);
          sb.append(' ');}out.println(sb);}void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0]
          .length);for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
  }