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
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
        while(t-->0) {
            char a[] = sc.next().toCharArray();
            char b[] = sc.next().toCharArray();
            boolean ok = false;
            for (int size = 0; size < b.length; size++) {
                int dp[][] = new int[size + 1][b.length - size + 1];
                int n = a.length;
                int next[][] = new int[n][26];
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < 26; j++) next[i][j] = n;
                }
                for (int i = 0; i < n; i++) next[i][a[i] - 'a'] = i;
 
                for (int j = 0; j < 26; j++) {
                    int first = n;
                    for (int i = n - 1; i >= 0; i--) {
                        first = Math.min(first, next[i][j]);
                        next[i][j] = first;
                    }
                }
 
                for (int i = 0; i <= size; i++) {
                    for (int j = 0; j <= b.length - size; j++) {
                        int ans = (int) (1e9+7);
                        if (i != 0 && dp[i - 1][j] + 1 < n) {
                            ans = next[dp[i - 1][j] + 1][b[i - 1] - 'a'];
                        }
                        if (j != 0 && dp[i][j - 1] + 1 < n) {
                            ans = Math.min(ans, next[dp[i][j - 1] + 1][b[size + j - 1] - 'a']);
                        }
                        if (i == 0 && j == 0) ans = -1;
 
                        dp[i][j] = ans;
                    }
                }
                if (dp[size][b.length - size] < n) ok = true;
            }
            System.out.println(ok ? "YES" : "NO");
	    }
    }
}
