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
#include <bits/stdc++.h>
using namespace std;
int T, n, m;
int f[405][405];
char s[405];
char t[405];
void work() {
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int k = 1; k <= m; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= k; j++) {
        f[i][j] = f[i - 1][j];
        if (j && f[i - 1][j - 1] >= 0 && s[i] == t[j]) {
          f[i][j] = max(f[i][j], f[i - 1][j - 1]);
        }
        if (f[i - 1][j] >= 0 && f[i - 1][j] + 1 <= m - k &&
            s[i] == t[k + f[i - 1][j] + 1]) {
          f[i][j] = max(f[i][j], f[i - 1][j] + 1);
        }
      }
    }
    if (f[n][k] == m - k) {
      printf("YES\n");
      return;
    }
  }
  printf("NO\n");
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1), m = strlen(t + 1);
    work();
  }
}
