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 N = 4e2 + 5;
char a[N], b[N];
long long dp[N][N];
int main() {
  long long n, m, i, j, k, ans, num, sum, t, len1, len2, sym;
  scanf("%lld", &t);
  while (t--) {
    scanf("%s", a + 1);
    scanf("%s", b + 1);
    len1 = strlen(a + 1);
    len2 = strlen(b + 1);
    sym = 0;
    for (k = 1; k <= len2; k++) {
      for (i = 1; i <= len1; i++)
        for (j = 0; j <= min(k - 1, i); j++) {
          dp[i][j] = (i - 1 >= j ? dp[i - 1][j] : -1e18);
          if (a[i] == b[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          if (i - 1 >= j && dp[i - 1][j] >= 0 && a[i] == b[k + dp[i - 1][j]])
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
        }
      if (dp[len1][k - 1] >= len2 - k + 1) {
        sym = 1;
        break;
      }
    }
    if (sym)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
