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 = int(410);
char s[N], r[N], ss[N], rr[N];
int dp[N][N];
const int INF = 100010;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("\n%s", ss);
    scanf("\n%s", rr);
    int n = strlen(ss);
    int m = strlen(rr);
    for (int i = 1; i <= n; i++) {
      s[i] = ss[i - 1];
    }
    for (int i = 1; i <= m; i++) {
      r[i] = rr[i - 1];
    }
    bool ok = false;
    for (int d = 0; d <= m; d++) {
      for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
          dp[i][j] = -INF;
        }
      }
      dp[0][0] = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= d; j++) {
          dp[i][j] = dp[i - 1][j];
          if (j > 0 && s[i] == r[j]) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          }
          int sz = dp[i - 1][j];
          if (sz >= 0 && d + sz + 1 <= m && s[i] == r[d + sz + 1]) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
          }
        }
        if (dp[i][d] == m - d) {
          ok = true;
          break;
        }
      }
      if (ok) break;
    }
    if (ok) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
