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[405][405];
string s, t;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    int n = s.size(), m = t.size();
    int flag = 0;
    for (int k = 1; k <= m; k++) {
      for (int i = 0; i <= n; i++)
        for (int j = 0; j <= m; j++) dp[i][j] = -0x3f3f3f3f;
      dp[0][0] = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= k; j++) {
          dp[i][j] = max(dp[i][j], dp[i - 1][j]);
          if (dp[i - 1][j] - j + k >= 0 && dp[i - 1][j] - j + k < m)
            dp[i][j] = max(
                dp[i][j], dp[i - 1][j] + (s[i - 1] == t[dp[i - 1][j] - j + k]));
          if (j >= 1 && s[i - 1] == t[j - 1])
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
        }
      }
      if (dp[n][k] == m) flag = 1;
    }
    if (flag)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}
