You are given a string s. You can build new string p from s using the following operation no more than two times: 

  1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|; 
  2. erase the chosen subsequence from s (s can become empty); 
  3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}). 



Of course, initially the string p is empty. 

For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.

Can you build a given string t using the algorithm above?

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.

The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.

It's guaranteed that the total length of strings s doesn't exceed 400.

Output

Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.

Example

Input


4
ababcd
abcba
a
b
defi
fed
xyz
x


Output


YES
NO
NO
YES
//Erase Subsequences
import java.io.*;
import java.util.*;

public class E1303{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 0; q < t; q++){
      
         char[] a = f.readLine().toCharArray();
         char[] b = f.readLine().toCharArray();
         int n = a.length;
         int nb = b.length;
         
         boolean found = false;
         
         for(int s = 1; s <= nb; s++){
            //split, first s points are in first part, next t-s are in second part.
            
            int[][] dp = new int[n+1][s+1];
            
            for(int k = 0; k <= n; k++){
               Arrays.fill(dp[k],-1);
            }
            
            dp[0][0] = 0;
            
            for(int k = 0; k < n; k++){
               for(int j = 0; j <= s; j++){
                  if(dp[k][j] == -1) continue;
                  if(j < s && a[k] == b[j]){
                     dp[k+1][j+1] = Math.max(dp[k+1][j+1],dp[k][j]);
                  }
                  if(s+dp[k][j] < nb && a[k] == b[s+dp[k][j]]){
                     dp[k+1][j] = Math.max(dp[k+1][j],dp[k][j]+1);
                  }
                  dp[k+1][j] = Math.max(dp[k+1][j],dp[k][j]);
               }
            }
            
            if(dp[n][s] == nb-s){
               found = true;
               break;
            }
         }
         
         if(found){
            out.println("YES");
         } else {
            out.println("NO");
         }
      }
   
      
      
      
      
      
      out.close();
   }
   
      
}