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.*;
import java.util.*;

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

        String next() throws Exception
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine()) ;
            return st.nextToken() ;
        }
        int nextInt() throws Exception { return Integer.parseInt(next()); }

    }
}