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 maxn = 410;
string S, T;
int t;
int nxt[maxn][30];
int memo[maxn][maxn];
int main() {
  cin >> t;
  while (t--) {
    cin >> S >> T;
    int n = S.size(), m = T.size();
    int last[30];
    for (int i = 0; i < 30; ++i) last[i] = n;
    for (int i = S.size() - 1; i >= 0; --i) {
      last[S[i] - 'a'] = i;
      for (int j = 0; j < 30; ++j) {
        nxt[i][j] = last[j];
      }
    }
    bool fun = 0;
    for (int k = 0; k <= m; ++k) {
      string A = T.substr(0, k), B = T.substr(k, m - k);
      int p = k, q = m - k;
      for (int i = 0; i <= p; ++i) {
        for (int j = 0; j <= q; ++j) {
          memo[i][j] = 1e9;
        }
      }
      memo[0][0] = 0;
      for (int i = 0; i <= p; ++i) {
        for (int j = 0; j <= q; ++j) {
          if (i > 0 and memo[i - 1][j] < n) {
            memo[i][j] =
                min(memo[i][j], nxt[memo[i - 1][j]][A[i - 1] - 'a'] + 1);
          }
          if (j > 0 and memo[i][j - 1] < n) {
            memo[i][j] =
                min(memo[i][j], nxt[memo[i][j - 1]][B[j - 1] - 'a'] + 1);
          }
        }
      }
      if (memo[p][q] <= n) {
        fun = 1;
        break;
      }
    }
    puts(fun ? "YES" : "NO");
  }
  return 0;
}
