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>
int t;
std::string a, b;
std::vector<std::vector<int>> dp;
void solve() {
  std::cin >> a >> b;
  int n = a.length(), m = b.length();
  for (int f = 0; f < m; ++f) {
    dp.assign(n + 1, std::vector<int>(f + 1, -1));
    dp[0][0] = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j <= f; ++j) {
        if (dp[i][j] == -1) continue;
        dp[i + 1][j] =
            std::max(dp[i + 1][j],
                     dp[i][j] + (dp[i][j] < m - f && a[i] == b[dp[i][j] + f]));
        if (j < f && a[i] == b[j])
          dp[i + 1][j + 1] = std::max(dp[i + 1][j + 1], dp[i][j]);
      }
    }
    if (dp[n][f] == m - f) {
      std::cout << "YES\n";
      return;
    }
  }
  std::cout << "NO\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> t;
  while (t--) solve();
  return 0;
}
