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 EraseSubsequences {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        for(int x = 0; x < t; x++){
            String s1 = f.readLine();
            String s2 = f.readLine();
            boolean found = false;
            for(int z = 0; z < s2.length(); z++){
                String sub1 = s2.substring(0, z+1);
                String sub2 = s2.substring(z+1);
                int[][] dp = new int[s1.length()+1][sub1.length()+1];
                for(int i = 0; i < s1.length()+1; i++){
                    Arrays.fill(dp[i], -1);
                }
                dp[0][0] = 0;
                for(int i = 0; i < s1.length(); i++){
                    for(int j = 0; j <= sub1.length(); j++){
                        if(j != sub1.length() && s1.charAt(i) == sub1.charAt(j)){
                            dp[i+1][j+1] = Math.max(dp[i+1][j+1], dp[i][j]);
                        }
                        if(dp[i][j] != -1 && dp[i][j] < sub2.length() && s1.charAt(i) == sub2.charAt(dp[i][j])){
                            dp[i+1][j] = Math.max(dp[i+1][j], dp[i][j]+1);
                        }
                        dp[i+1][j] = Math.max(dp[i+1][j], dp[i][j]);
                    }
                }
                for(int i = 0; i <= s1.length(); i++){
                    if(dp[i][sub1.length()] == sub2.length()){
                        found = true;
                    }
                }
                if(found) break;
            }
            if(found){
                out.println("YES");
            }else{
                out.println("NO");
            }
        }
        out.close();
    }
}
