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 f[409][409];
char s[409], t[409];
int main() {
  int T, i, j, k, n, m;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1), n = strlen(s + 1), m = strlen(t + 1);
    for (i = 1; i <= m; ++i) {
      for (j = 1, k = 1; j <= n; ++j) {
        while (j <= n && s[j] != t[k]) ++j;
        if (j > n) {
          puts("NO");
          goto ggg;
        }
        ++k;
        if (k > i) break;
      }
      for (j = 0; j <= n; ++j) {
        for (k = 0; k <= i; ++k) {
          f[j][k] = -1e9;
        }
      }
      f[0][0] = 0;
      for (j = 1; j <= n; ++j) {
        for (k = 0; k <= i; ++k) {
          f[j][k] = f[j - 1][k];
          if (f[j - 1][k] >= 0 && s[j] == t[i + f[j - 1][k] + 1]) ++f[j][k];
          if (s[j] == t[k] && k > 0 && f[j][k] < f[j - 1][k - 1])
            f[j][k] = f[j - 1][k - 1];
        }
      }
      if (f[n][i] == m - i) {
        puts("YES");
        goto ggg;
      }
    }
    puts("NO");
  ggg:;
  }
  return 0;
}
