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 = 405;
int n, m;
char S[MAXN], T[MAXN];
bitset<MAXN> f[MAXN];
int main() {
  int Ti;
  scanf("%d", &Ti);
  while (Ti--) {
    scanf("%s%s", S + 1, T + 1);
    n = strlen(S + 1), m = strlen(T + 1);
    bool ans = false;
    for (int p = 1; p <= m && !ans; ++p) {
      for (int i = 0; i <= p; ++i) f[i].reset();
      f[0][0] = 1;
      for (int i = 1; i <= n; ++i) {
        bitset<MAXN> d;
        for (int j = 1; j <= m - p; ++j) d[j - 1] = S[i] == T[p + j];
        for (int j = p; j >= 0; --j) {
          f[j] |= (f[j] & d) << 1;
          if (j > 0 && S[i] == T[j]) f[j] |= f[j - 1];
        }
      }
      ans = f[p][m - p];
    }
    puts(ans ? "YES" : "NO");
  }
  return 0;
}
