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[410][410], n, m, l;
char s[410], a[410], b[410], t[410];
int nex[410][30];
int solve() {
  int i, j, k;
  for (i = 0; i <= n; i++)
    for (j = 0; j <= m; j++) f[i][j] = l + 1;
  f[0][0] = 0;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      if (i != 0 && f[i - 1][j] != l + 1)
        f[i][j] = min(f[i][j], nex[f[i - 1][j]][a[i] - 'a']);
      if (j != 0 && f[i][j - 1] != l + 1)
        f[i][j] = min(f[i][j], nex[f[i][j - 1]][b[j] - 'a']);
    }
  }
  return f[n][m] != l + 1;
}
int main() {
  int i, T, j;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    l = strlen(s + 1);
    memset(nex, 0, sizeof(nex));
    for (i = l; i >= 0; i--) {
      if (l == i)
        for (j = 0; j < 26; j++) nex[i][j] = l + 1;
      else {
        for (j = 0; j < 26; j++) nex[i][j] = nex[i + 1][j];
        nex[i][s[i + 1] - 'a'] = i + 1;
      }
    }
    int ans = 0;
    for (i = 1; t[i]; i++) {
      n = m = 0;
      for (j = 1; j <= i; j++) a[++n] = t[j];
      for (j = i + 1; t[j]; j++) b[++m] = t[j];
      if (solve()) {
        ans = 1;
        break;
      }
    }
    if (ans)
      printf("YES\n");
    else
      printf("NO\n");
  }
}
