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;
void read(int &x) {
  char ch = getchar();
  x = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
}
const int N = 410;
int n, m, f[N][N];
char a[N], b[N];
signed main() {
  int T;
  read(T);
  while (T--) {
    scanf("%s %s", a + 1, b + 1);
    n = strlen(a + 1), m = strlen(b + 1);
    int tag = 0;
    for (int mid = 1; mid <= m; ++mid) {
      for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= mid; ++j) f[i][j] = -1;
      f[0][0] = 0;
      for (int i = 1; i <= n; ++i)
        for (int j = 0; j <= mid; ++j) {
          if (f[i - 1][j] >= 0)
            f[i][j] = f[i - 1][j] + (a[i] == b[mid + 1 + f[i - 1][j]]);
          if (a[i] == b[j] && j)
            (f[i][j] = f[i][j] > f[i - 1][j - 1] ? f[i][j] : f[i - 1][j - 1]);
        }
      if (f[n][mid] == m - mid) {
        tag = 1;
        puts("YES");
        break;
      }
    }
    if (!tag) puts("NO");
  }
  return 0;
}
