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 a[405];
char b[405];
int dp[405][405];
void work() {
  scanf("%s", a + 1);
  scanf("%s", b + 1);
  int l1 = strlen(a + 1);
  int l2 = strlen(b + 1);
  for (int i = 1; i <= l2; i++) {
    for (int j = 0; j <= l1; j++) {
      for (int k = 0; k <= min(i, j); k++) {
        dp[j][k] = -1;
      }
    }
    dp[0][0] = i;
    for (int j = 0; j < l1; j++) {
      for (int k = 0; k <= min(i, j); k++) {
        if (k < i && a[j + 1] == b[k + 1])
          dp[j + 1][k + 1] = max(dp[j + 1][k + 1], dp[j][k]);
        if (dp[j][k] < l2 && a[j + 1] == b[dp[j][k] + 1])
          dp[j + 1][k] = max(dp[j + 1][k], dp[j][k] + 1);
        dp[j + 1][k] = max(dp[j + 1][k], dp[j][k]);
      }
    }
    if (dp[l1][i] == l2) {
      printf("YES\n");
      return;
    }
  }
  printf("NO\n");
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    work();
  }
}
