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 = 4e2 + 10;
int T;
string s, t;
int dp[maxn][maxn];
int main() {
  cin >> T;
  while (T--) {
    cin >> s >> t;
    int n = s.size(), m = t.size();
    s = '#' + s, t = '#' + t;
    bool flag = 0;
    for (int d = 0; d <= m; d++) {
      memset(dp, 0xcf, sizeof(dp));
      dp[0][0] = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= min(d, i); j++) {
          if (s[i] == t[j]) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          }
        }
        for (int j = 0; j <= min(d, i); j++) {
          if (dp[i - 1][j] >= 0) {
            dp[i][j] =
                max(dp[i][j], dp[i - 1][j] + (s[i] == t[d + 1 + dp[i - 1][j]]));
          }
        }
      }
      if (dp[n][d] == m - d) {
        flag = 1;
        break;
      }
    }
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
