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;
mt19937 mt_rand(time(0));
const int N = 405;
int T;
char s[N], t[N];
int dp[N][N];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    int n = 1, m = 1;
    while (s[n]) n++;
    while (t[m]) m++;
    bool ok = 0;
    for (int p = 1; p < m; p++) {
      dp[0][0] = p;
      for (int i = 1; i < n; i++) dp[0][i] = -1;
      for (int i = 1; i < n; i++)
        for (int j = 0; j <= p; j++) {
          dp[i][j] = dp[i - 1][j];
          if (j != 0 && s[i] == t[j])
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          int val = dp[i - 1][j];
          if (val != -1 && val != m - 1 && t[val + 1] == s[i])
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
        }
      if (dp[n - 1][p] == m - 1) ok = 1;
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
