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 main() {
  int tttt;
  cin >> tttt;
  for (int loop = 0; loop < (tttt); ++loop) {
    string s, t;
    cin >> s >> t;
    int n = s.size();
    int m = t.size();
    int flag = 0;
    int now = 0;
    for (int i = 0; i < (n); ++i) {
      if (now == m) break;
      if (s[i] == t[now]) now++;
    }
    if (now == m) flag = 1;
    for (int i = 1; i < m; ++i) {
      if (flag) break;
      int dp[n][i + 1];
      for (int j = 0; j < (n); ++j)
        for (int k = 0; k < (i + 1); ++k) dp[j][k] = -1;
      for (int j = 0; j < (n); ++j) {
        for (int k = 0; k < (i + 1); ++k) {
          if (k > j + 1) break;
          if (j == 0) {
            if (k == 0) {
              dp[j][k] = i;
              if (s[j] == t[i]) dp[j][k] = i + 1;
            } else {
              if (s[j] == t[0]) dp[j][k] = i;
            }
          } else {
            dp[j][k] = dp[j - 1][k];
            if (dp[j][k] != -1 && dp[j][k] < m) {
              if (s[j] == t[dp[j][k]]) dp[j][k]++;
            }
            if (k > 0) {
              if (dp[j - 1][k - 1] != -1 && s[j] == t[k - 1]) {
                dp[j][k] = max(dp[j][k], dp[j - 1][k - 1]);
              }
            }
          }
        }
      }
      if (dp[n - 1][i] == m) flag = 1;
    }
    if (flag) {
      puts("YES");
    } else {
      puts("NO");
    };
  }
  return 0;
}
