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() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return x * f;
}
long long k, m, n, net[410][27], dp[410][410];
char c[410], s[410];
signed main() {
  long long t = read();
  while (t--) {
    memset(net, 0, sizeof(net));
    scanf("%s", c + 1);
    n = strlen(c + 1);
    for (register long long i = 1; i <= n; i++) {
      net[i - 1][c[i] - 'a' + 1] = i;
    }
    for (register long long i = 1; i <= 26; i++) {
      for (register long long j = n - 1; j >= 0; j--) {
        if (net[j][i]) continue;
        net[j][i] = net[j + 1][i];
      }
    }
    scanf("%s", s + 1);
    m = strlen(s + 1);
    long long ans = 1e9;
    for (register long long now = 1; now <= m; now++) {
      memset(dp, 0x3f, sizeof(dp));
      dp[0][0] = 0;
      for (register long long i = 0; i <= now; i++) {
        for (register long long j = 0; j <= m - now; j++) {
          if (i != 0 && dp[i - 1][j] <= n &&
              net[dp[i - 1][j]][s[i] - 'a' + 1]) {
            dp[i][j] = min(dp[i][j], net[dp[i - 1][j]][s[i] - 'a' + 1]);
          }
          if (j != 0 && dp[i][j - 1] <= n &&
              net[dp[i][j - 1]][s[now + j] - 'a' + 1]) {
            dp[i][j] = min(dp[i][j], net[dp[i][j - 1]][s[now + j] - 'a' + 1]);
          }
        }
      }
      ans = min(ans, dp[now][m - now]);
    }
    if (ans <= n)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
