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 dp[500][500];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    int n = s.length(), m = t.length();
    bool is = 0;
    for (int div = 1; div <= m; div++) {
      for (int i = 0; i < 500; i++)
        for (int j = 0; j < 500; j++) dp[i][j] = -1;
      for (int i = 0; i < 500; i++) dp[i][0] = 0;
      if (s[0] == t[div]) dp[0][0] = 1;
      if (s[0] == t[0]) dp[0][1] = 0;
      if (m == 1)
        if (s[0] == t[0]) is = 1;
      for (int i = 1; i < n; i++) {
        for (int reach = 0; reach <= div; reach++) {
          dp[i][reach] = dp[i - 1][reach];
          if (reach > 0)
            if (s[i] == t[reach - 1])
              if (dp[i - 1][reach - 1] != -1)
                dp[i][reach] = max(dp[i][reach], dp[i - 1][reach - 1]);
          if (dp[i - 1][reach] != -1)
            if (s[i] == t[dp[i - 1][reach] + div])
              dp[i][reach] = max(dp[i][reach], dp[i - 1][reach] + 1);
          if (reach == div && dp[i][reach] == m - div) is = 1;
        }
      }
    }
    cout << (is ? "YES\n" : "NO\n");
  }
  return 0;
}
