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;
long long Q;
string s, t;
short dp[405][405][405];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    cin >> s >> t;
    long long S = s.size(), T = t.size();
    s = "#" + s, t = "#" + t;
    for (int i = 0; i <= S; i++) {
      for (int j = 0; j <= T; j++) {
        for (int k = 0; k <= T; k++) {
          dp[i][j][k] = 1000;
        }
      }
    }
    for (int i = 0; i <= T; i++) dp[0][0][i] = i;
    for (int i = 0; i < S; i++) {
      for (int j = 0; j <= T; j++) {
        for (int k = 0; k <= T; k++) {
          if (s[i + 1] == t[j + 1] && j + 1 <= T) {
            dp[i + 1][j + 1][k] = min(dp[i + 1][j + 1][k], dp[i][j][k]);
          }
          if (s[i + 1] == t[k + 1] && k + 1 <= T) {
            dp[i + 1][j][k + 1] = min(dp[i + 1][j][k + 1], dp[i][j][k]);
          }
          dp[i + 1][j][k] = min(dp[i + 1][j][k], dp[i][j][k]);
        }
      }
    }
    for (int i = 0; i <= S; i++) {
      for (int j = 0; j <= T; j++) {
        if (dp[i][j][T] <= j) {
          cout << "YES" << endl;
          goto end;
        }
      }
    }
    cout << "NO" << endl;
  end:;
  }
  flush(cout);
  return 0;
}
