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 INF = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    int n = s.size(), m = t.size();
    vector<vector<int>> nxt(n, vector<int>(26, n));
    nxt[n - 1][s[n - 1] - 'a'] = n - 1;
    for (int i = n - 2; i >= 0; i--) {
      for (int j = 0; j < 26; j++) {
        nxt[i][j] = nxt[i + 1][j];
      }
      nxt[i][s[i] - 'a'] = i;
    }
    bool ans = false;
    for (int i = 0; i <= m - 1; i++) {
      vector<vector<int>> dp(i + 1, vector<int>(m - i + 1, INF));
      dp[0][0] = 0;
      for (int j = 0; j <= i; j++) {
        for (int k = 0; k <= m - i; k++) {
          if (dp[j][k] < n) {
            if (j < i) {
              dp[j + 1][k] = min(dp[j + 1][k], nxt[dp[j][k]][t[j] - 'a'] + 1);
            }
            if (k < m - i) {
              dp[j][k + 1] =
                  min(dp[j][k + 1], nxt[dp[j][k]][t[k + i] - 'a'] + 1);
            }
          }
        }
      }
      if (dp[i][m - i] <= n) {
        ans = true;
      }
    }
    if (ans)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
