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 = 410;
int tu, n, m, dp[MAXN][MAXN], nx[MAXN][27];
char s[MAXN], t[MAXN];
bool chk(int x) {
  for (int i = 0; i <= x - 1; i++) {
    for (int j = x - 1; j <= m; j++) {
      if (i == 0 && j == x - 1) {
        dp[i][j] = 0;
        continue;
      }
      if (i == 0) {
        dp[i][j] = nx[dp[i][j - 1]][t[j] - 'a' + 1];
        continue;
      }
      if (j == x - 1) {
        dp[i][j] = nx[dp[i - 1][j]][t[i] - 'a' + 1];
        continue;
      }
      dp[i][j] = min(nx[dp[i - 1][j]][t[i] - 'a' + 1],
                     nx[dp[i][j - 1]][t[j] - 'a' + 1]);
    }
  }
  return (dp[x - 1][m] <= n);
}
int main() {
  scanf("%d", &tu);
  for (int ii = 1; ii <= tu; ii++) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1), m = strlen(t + 1);
    memset(nx, 0, sizeof(nx));
    for (int j = 1; j <= 26; j++) {
      nx[n][j] = nx[n + 1][j] = n + 1;
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 1; j <= 26; j++) {
        nx[i][j] = nx[i + 1][j];
      }
      nx[i][s[i + 1] - 'a' + 1] = i + 1;
    }
    int flg = 0;
    for (int i = 1; i <= m + 1; i++) {
      if (chk(i)) {
        printf("YES\n");
        flg = 1;
        break;
      }
    }
    if (!flg) {
      printf("NO\n");
    }
  }
  return 0;
}
