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 long long N = 410;
long long T, n, l1, l2, k, f[N][N], cnt;
bool flag;
string s, t, t1, t2;
signed main() {
  scanf("%lld", &T);
  while (T--) {
    cin >> s >> t, flag = cnt = 0, n = s.size();
    for (long long i = 0; i < n; i++) {
      if (s[i] == t[cnt]) ++cnt;
      if (cnt == t.size()) {
        puts("YES"), flag = 1;
        break;
      }
    }
    if (flag) continue;
    for (long long ys = 1; ys < t.size(); ys++) {
      t1 = t.substr(0, ys), t2 = t.substr(ys), l1 = t1.size(), l2 = t2.size();
      memset(f, -1, sizeof(f)), f[0][0] = 0;
      for (long long i = 0; i < n; i++)
        for (long long j = 0; j <= l1; j++)
          if ((k = f[i][j]) >= 0) {
            if (j < l1 && s[i] == t1[j])
              f[i + 1][j + 1] = max(k, f[i + 1][j + 1]);
            if (k < l2 && s[i] == t2[k]) f[i + 1][j] = max(k + 1, f[i + 1][j]);
            f[i + 1][j] = max(k, f[i + 1][j]);
          }
      if (f[n][l1] == l2) {
        puts("YES"), flag = 1;
        break;
      }
    }
    if (!flag) puts("NO");
  }
  return 0;
}
