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 T, len1, len2;
int nex[maxn][31];
int dp[maxn][maxn];
char s[maxn], t[maxn];
void ready(int n) {
  for (int i = 1; i <= 26; i++) nex[n][i] = n + 1;
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= 26; j++) nex[i - 1][j] = nex[i][j];
    nex[i - 1][s[i] - 'a' + 1] = i;
  }
}
bool check(int n) {
  int m = len2 - n;
  dp[0][0] = 0;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      if (!i && !j) continue;
      dp[i][j] = len1 + 1;
      if (i && dp[i - 1][j] < len1)
        dp[i][j] = min(dp[i][j], nex[dp[i - 1][j]][t[i] - 'a' + 1]);
      if (j && dp[i][j - 1] < len1)
        dp[i][j] = min(dp[i][j], nex[dp[i][j - 1]][t[n + j] - 'a' + 1]);
    }
  return dp[n][m] <= len1;
}
void solve() {
  scanf("%s%s", s + 1, t + 1);
  len1 = strlen(s + 1);
  len2 = strlen(t + 1);
  ready(len1);
  for (int i = 1; i <= len2; i++)
    if (check(i)) {
      printf("YES\n");
      return;
    }
  printf("NO\n");
  return;
}
int main() {
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
