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
import java.io.*;

public class Main
{
    static int dp(int [][][] memo , char [] s , char [] t , int index , int i , int j , int endI , int endJ)
    {
        if(i > endI || j > endJ) return -(int)1e9 ;
        if(index == s.length)  return i == endI && j == endJ ? 0 : -(int)1e9 ;
        if(memo[index][i][j] != -1)return memo[index][i][j];
        int ans = dp(memo , s , t , index + 1 , i , j , endI , endJ);
        if(i < endI && s[index] == t[i]) ans = Math.max(ans , 1 + dp(memo , s , t ,index + 1 , i + 1 , j , endI , endJ));
        if(j < endJ && s[index] == t[j]) ans = Math.max(ans , 1 + dp(memo , s , t , index + 1 , i , j + 1 , endI , endJ)) ;
        return memo[index][i][j] = ans ;
    }
    public static void main(String[] args) throws Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out) ;
        int TC = Integer.parseInt(br.readLine()) ;
        int memo[][][] = new int [401][401][401];
        outer:
        while (TC -->0)
        {
            char [] s = br.readLine().toCharArray() , t = br.readLine().toCharArray();
            for(int i = 0 ; i <= s.length ; i ++)
                for(int j = 0 ; j <= t.length ; j++)
                    for(int k = 0 ; k <= t.length ; k++)
                        memo[i][j][k] = -1 ;
            for(int i = 1 ; i <= t.length ; i++)
                if(dp(memo , s , t , 0 , 0 , i , i , t.length) == t.length)
                {
                    out.println("YES");
                    continue outer;
                }
            out.println("NO");
        }
        out.flush();
    }
  }