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 M = 405;
int dp[M][M], n, m;
char S[M], T[M];
bool solve() {
  scanf("%s", S + 1);
  scanf("%s", T + 1);
  int n = strlen(S + 1);
  int m = strlen(T + 1);
  int s1 = 0;
  for (int s2 = 1; s2 <= m; ++s2) {
    memset(dp, -1, sizeof(dp));
    int l1 = s2 - s1, l2 = m - s2;
    dp[0][0] = 0;
    for (int i = 1; i <= n; ++i)
      for (int a = 0; a <= l1; ++a)
        if (~dp[i - 1][a]) {
          int b = dp[i - 1][a];
          if (a < l1 && S[i] == T[s1 + a + 1])
            dp[i][a + 1] = max(dp[i][a + 1], b);
          if (b < l2 && S[i] == T[s2 + b + 1]) dp[i][a] = max(dp[i][a], b + 1);
          dp[i][a] = max(dp[i][a], b);
        }
    if (dp[n][l1] == l2) return 1;
  }
  return 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    if (solve())
      printf("YES\n");
    else
      printf("NO\n");
  }
}
