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[405], t[405];
int f[2][405][405];
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    int lens = strlen(s + 1), lent = strlen(t + 1);
    int cur = 0, pre = 1;
    memset(f, -63, sizeof f);
    for (int i = 0; i < lent; ++i) f[cur][0][i] = 0;
    for (int i = 1; i <= lens; ++i) {
      swap(cur, pre);
      for (int j = 0; j <= lent; ++j)
        for (int k = j; k <= lent; ++k) {
          f[cur][j][k] = f[pre][j][k];
          if (j && s[i] == t[j])
            f[cur][j][k] = max(f[cur][j][k], f[pre][j - 1][k]);
          if (k && s[i] == t[k])
            f[cur][j][k] = max(f[cur][j][k], f[pre][j][k - 1] + 1);
        }
    }
    bool flag = false;
    for (int i = 0; i <= lent; ++i)
      if (lent - f[cur][i][lent] <= i) flag = true;
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
