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 MAXN = 400;
int n, m, dp[MAXN + 5][MAXN + 5];
char s[MAXN + 5], t[MAXN + 5];
inline void ckmax(int &x, int y) { x = x > y ? x : y; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    bool ok = 0;
    for (int i = 1; i <= m; ++i) {
      memset(dp, -1, sizeof(dp));
      dp[0][0] = 0;
      for (int j = 1; j <= n; ++j) {
        for (int k = 0; k <= i; ++k) {
          if (dp[j - 1][k] != -1) {
            ckmax(dp[j][k], dp[j - 1][k] + (i + dp[j - 1][k] + 1 <= m &&
                                            s[j] == t[i + dp[j - 1][k] + 1]));
          }
          if (k >= 1 && s[j] == t[k]) {
            ckmax(dp[j][k], dp[j - 1][k - 1]);
          }
        }
      }
      if (dp[n][i] == m - i) {
        ok = 1;
        break;
      }
    }
    if (ok)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
