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.util.Arrays;
import java.util.Scanner;

public class P1303E2 {

	private static Scanner in;
	private static String s, t;
	private static int n, m;
	private static int[][] next;
	
	public static void main(String[] args) {
		in = new Scanner(System.in);
		int q = in.nextInt();
		next:
		while(q-->0) {
			s = in.next();
			t = in.next();
			n = s.length();
			m = t.length();
			next = new int[n][26];
			for(int i=0; i<n; i++) {
				Arrays.fill(next[i], n);
			}
			for(int i=0; i<n; i++) {
				for(int j=n-1; j>=i; j--) {
					next[i][s.charAt(j)-'a'] = j;
				}
			}
			
			int[][] dp = new int[m+1][m+1];
			
			for(int p=0; p<m; p++) {
				for(int i=0; i<dp.length; i++) {
					Arrays.fill(dp[i], n+1);
				}
				dp[0][0] = 0;
				String a = t.substring(0, p+1);
				String b = t.substring(p+1, m);
				for(int i=0; i<=a.length(); i++) {
					for(int j=0; j<=b.length(); j++) {
						int len = dp[i][j];
						if (len>=n) continue;
						if (i<a.length() && next[len][a.charAt(i)-'a']<n) {
							dp[i+1][j] = Math.min(dp[i+1][j], next[len][a.charAt(i)-'a']+1);
						}
						if (j<b.length() && next[len][b.charAt(j)-'a']<n) {
							dp[i][j+1] = Math.min(dp[i][j+1], next[len][b.charAt(j)-'a']+1);
						}
					}
				}
//				System.out.println("PARTITION: "+p);
//				for(int i=0; i<=a.length(); i++) {
//					for(int j=0; j<=b.length(); j++) {
//						System.out.print(dp[i][j]+" ");
//					}
//					System.out.println();
//				}
				if (dp[a.length()][b.length()]<=n) {
					System.out.println("YES");
					continue next;
				}
				
			}
			System.out.println("NO");
		}
	}
}
