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;
int t;
int n, m;
string A, B;
int f[405][405];
int main() {
  int i, j, k;
  cin >> t;
  while (t--) {
    cin >> A >> B;
    n = A.length(), m = B.length();
    A = "0" + A;
    B = "0" + B;
    i = 1, j = 1;
    for (i = 1; i <= n; i++) {
      if (A[i] == B[j]) {
        j++;
        if (j == m + 1) break;
      }
    }
    if (j == m + 1) {
      cout << "YES" << endl;
      continue;
    }
    bool flag = false;
    for (k = 1; k < n; k++) {
      memset(f, 0, sizeof(f));
      f[0][0] = k;
      for (i = 0; i < n; i++) {
        for (j = 0; j <= k; j++) {
          if (f[i][j] == 0) continue;
          f[i + 1][j] = max(f[i + 1][j], f[i][j]);
          if (j < k and A[i + 1] == B[j + 1]) {
            f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
          }
          if (f[i][j] < m and A[i + 1] == B[f[i][j] + 1]) {
            f[i + 1][j] = max(f[i + 1][j], f[i][j] + 1);
          }
        }
      }
      if (f[n][k] == m) {
        flag = true;
        break;
      }
    }
    if (flag) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
