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;
long long read() {
  char ch = getchar();
  long long x = 0, ff = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -ff;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * ff;
}
void write(long long aa) {
  if (aa < 0) putchar('-'), aa = -aa;
  if (aa > 9) write(aa / 10);
  putchar(aa % 10 + '0');
  return;
}
long long T;
long long n, m;
long long nx[405][28], f[405][405];
char s[405], t[405];
int main() {
  T = read();
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    for (int i = 0; i < 26; ++i) nx[n][i] = n + 1, nx[n + 1][i] = n + 1;
    for (int i = n; i; --i) {
      for (int j = 0; j < 26; ++j) nx[i - 1][j] = nx[i][j];
      nx[i - 1][s[i] - 'a'] = i;
    }
    int FF = 0;
    for (int i = 0; i <= m; ++i) {
      f[0][0] = 0;
      for (int j = 0; j <= i; ++j)
        for (int k = 0; k <= m - i; ++k)
          if (j || k) {
            f[j][k] = n + 1;
            if (j) f[j][k] = min(nx[f[j - 1][k]][t[j] - 'a'], f[j][k]);
            if (k) f[j][k] = min(nx[f[j][k - 1]][t[i + k] - 'a'], f[j][k]);
          }
      if (f[i][m - i] <= n) {
        puts("YES");
        FF = 1;
        break;
      }
    }
    if (!FF) puts("NO");
  }
  return 0;
}
