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;
char s[1000], t[1000];
int f[405], g[405];
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%s%s", s, t);
    int n = strlen(s);
    int m = strlen(t);
    bool ans = false;
    for (int p = 0; p < m; p++) {
      vector<char> v1, v2;
      for (int i = 0; i <= p; i++) v1.push_back(t[i]);
      for (int i = p + 1; i < m; i++) v2.push_back(t[i]);
      memset(f, 0xff, sizeof f);
      f[0] = 0;
      for (int i = 0; i < n; i++) {
        memcpy(g, f, sizeof g);
        for (int j = 0; j <= v1.size(); j++) {
          if (f[j] >= 0 && j < v1.size() && s[i] == v1[j])
            g[j + 1] = max(g[j + 1], f[j]);
          if (f[j] >= 0 && f[j] < v2.size() && s[i] == v2[f[j]])
            g[j] = max(g[j], f[j] + 1);
        }
        memcpy(f, g, sizeof f);
      }
      if (f[v1.size()] == v2.size()) {
        ans = true;
        break;
      }
    }
    puts(ans ? "YES" : "NO");
  }
  return 0;
}
