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;
vector<vector<int> > nxt, dp;
int main() {
  int T;
  cin >> T;
  while (T--) {
    string s, t, a = "", b = "";
    cin >> s >> t;
    int n = s.size(), m = t.size();
    nxt.assign(n + 1, vector<int>(26, INF));
    for (int i = n - 1; i >= 0; i--) {
      nxt[i] = nxt[i + 1];
      nxt[i][s[i] - 'a'] = i;
    }
    bool ok = false;
    for (int i = 0; i < m; i++) {
      a = t.substr(0, i), b = t.substr(i, m);
      int as = a.size(), bs = b.size();
      dp.assign(as + 1, vector<int>(bs + 1, INF));
      dp[0][0] = 0;
      for (int j = 0; j <= as; j++) {
        for (int k = 0; k <= bs; k++) {
          int len = dp[j][k];
          if (len > n) continue;
          if (j < as && nxt[len][a[j] - 'a'] < INF) {
            dp[j + 1][k] = min(dp[j + 1][k], nxt[len][a[j] - 'a'] + 1);
          }
          if (k < bs && nxt[len][b[k] - 'a'] < INF) {
            dp[j][k + 1] = min(dp[j][k + 1], nxt[len][b[k] - 'a'] + 1);
          }
        }
      }
      if (dp[as][bs] <= n) {
        ok = true;
        break;
      }
    }
    if (ok) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}
