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 = 410;
int T;
char s[N], t[N];
int n, m;
int nxt[N][30], pre[30];
int dp[N][N];
void get_dp(int p) {
  for (int i = 0; i <= p; i++) {
    for (int j = 0; j <= m - p; j++) {
      if (i == 0 && j == 0)
        dp[i][j] = 0;
      else if (i == 0)
        dp[i][j] = nxt[dp[i][j - 1]][t[j + p] - 'a'];
      else if (j == 0)
        dp[i][j] = nxt[dp[i - 1][j]][t[i] - 'a'];
      else
        dp[i][j] = min(nxt[dp[i][j - 1]][t[j + p] - 'a'],
                       nxt[dp[i - 1][j]][t[i] - 'a']);
    }
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    fill(nxt[405], nxt[405] + 26, 405);
    fill(pre, pre + 26, 405);
    for (int i = n; i >= 1; i--) {
      memcpy(nxt[i], pre, sizeof(pre));
      pre[s[i] - 'a'] = i;
    }
    memcpy(nxt[0], pre, sizeof(pre));
    bool ok = false;
    for (int i = 1; i <= m; i++) {
      get_dp(i);
      if (dp[i][m - i] != 405) ok = true;
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  getchar();
  getchar();
  return 0;
}
