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;
using LL = long long;
const int maxn = 405;
const int inf = 0x3f3f3f3f;
char s[maxn], t[maxn];
int dp[maxn][maxn];
int nxt[maxn][26];
int main() {
  int T;
  scanf("%d", &T);
  int i, j, k;
  while (T--) {
    memset(nxt, 0x3f, sizeof(nxt));
    scanf("%s%s", s + 1, t + 1);
    int len = strlen(t + 1);
    int mx = strlen(s + 1);
    for (i = mx; i >= 1; i--) {
      for (j = 0; j < 26; j++) {
        nxt[i][j] = nxt[i + 1][j];
      }
      nxt[i][s[i] - 'a'] = i;
    }
    int res;
    bool flag = 0;
    for (i = 1; i <= len; i++) {
      res = len - i;
      memset(dp, 0x3f, sizeof(dp));
      dp[0][0] = 0;
      for (j = 0; j <= i; j++) {
        for (k = 0; k <= res; k++) {
          if (j && dp[j - 1][k] < mx)
            dp[j][k] = min(dp[j][k], nxt[dp[j - 1][k] + 1][t[j] - 'a']);
          if (k && dp[j][k - 1] < mx)
            dp[j][k] = min(dp[j][k], nxt[dp[j][k - 1] + 1][t[k + i] - 'a']);
        }
      }
      if (dp[i][res] <= mx) {
        flag = 1;
        break;
      }
    }
    if (flag == 1)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
