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>
#pragma optimize GCC("O3")
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    string t;
    cin >> s >> t;
    int n = s.size();
    int m = t.size();
    vector<vector<int>> nxt(n, vector<int>(26, -1));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < nxt[i].size(); j++) {
        for (int k = i; k < n; k++) {
          if (s[k] == j + 'a') {
            nxt[i][j] = k;
            break;
          }
        }
      }
    }
    bool yes = 0;
    for (int r = 0; r <= m; r++) {
      string a, b;
      for (int i = 0; i < r; i++) a.push_back(t[i]);
      for (int i = r; i < m; i++) b.push_back(t[i]);
      vector<vector<int>> dp(n + 1, vector<int>(n + 1, 1079));
      dp[0][0] = -1;
      for (int i = 0; i <= a.size(); i++) {
        for (int j = 0; j <= b.size(); j++) {
          if (dp[i][j] > n - 2) continue;
          if (i < a.size() && nxt[dp[i][j] + 1][a[i] - 'a'] != -1)
            dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j] + 1][a[i] - 'a']);
          if (j < b.size() && nxt[dp[i][j] + 1][b[j] - 'a'] != -1)
            dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j] + 1][b[j] - 'a']);
        }
      }
      if (dp[a.size()][b.size()] != 1079) {
        yes = 1;
        break;
      }
    }
    if (yes)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
