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 CF1303E extends PrintWriter {
	CF1303E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1303E o = new CF1303E(); o.main(); o.flush();
	}

	static final int A = 26;
	void main() {
		int[] ii = new int[A];
		int t = sc.nextInt();
		while (t-- > 0) {
			byte[] ss = sc.next().getBytes();
			byte[] tt = sc.next().getBytes();
			int n = ss.length;
			int m = tt.length;
			for (int i = 0; i < n; i++)
				ss[i] -= 'a';
			for (int i = 0; i < m; i++)
				tt[i] -= 'a';
			int[][] next = new int[n][A];
			Arrays.fill(ii, n);
			for (int i = n - 1; i >= 0; i--) {
				ii[ss[i]] = i;
				for (int a = 0; a < A; a++)
					next[i][a] = ii[a];
			}
			int[][] dp = new int[m + 1][m + 1];
			boolean yes = false;
			for (int l = 0; l <= m; l++) {
				for (int i = 0; i <= l; i++)
					for (int j = l; j <= m; j++)
						dp[i][j] = n + 1;
				dp[0][l] = 0;
				for (int i = 0; i <= l; i++)
					for (int j = l; j <= m; j++) {
						int h = dp[i][j];
						if (h >= n)
							continue;
						if (i < l)
							dp[i + 1][j] = Math.min(dp[i + 1][j], next[h][tt[i]] + 1);
						if (j < m)
							dp[i][j + 1] = Math.min(dp[i][j + 1], next[h][tt[j]] + 1);
					}
				if (dp[l][m] <= n) {
					yes = true;
					break;
				}
			}
			println(yes ? "YES" : "NO");
		}
	}
}
