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;
const int N = 410;
int dp[N][N];
int Main() {
  string s, p;
  cin >> s >> p;
  int n = s.size(), m = p.size();
  for (int t = 0; t < m; t++) {
    memset(dp, -1, sizeof dp), dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= t; j++) {
        dp[i][j] = dp[i - 1][j] + (0 <= dp[i - 1][j] && dp[i - 1][j] < m - t
                                       ? s[i - 1] == p[t + dp[i - 1][j]]
                                       : 0);
        if (j > 0 && s[i - 1] == p[j - 1])
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
    if (dp[n][t] == m - t) return cout << "YES\n", 0;
  }
  return cout << "NO\n", 0;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    Main();
  }
}
