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;
long long mod = 1e9 + 7;
long long fast_exp(long long a, long long b) {
  if (b <= 0)
    return 1;
  else {
    long long res = 1;
    res = fast_exp(a, b / 2);
    res = (res * res) % mod;
    if (b % 2 == 1) res = (res * a) % mod;
    return res;
  }
}
const long long N = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    int n = s.size();
    int m = t.size();
    int split = 0;
    int chk = 0;
    for (split = 0; split < m; split++) {
      int dp[n][split + 5];
      for (int i = 0; i < n; i++)
        for (int j = 0; j < split + 5; j++) {
          dp[i][j] = -1;
          if (j == 0) dp[i][j] = 0;
        }
      if (s[0] == t[0]) dp[0][1] = 0;
      if (s[0] == t[split]) dp[0][0] = 1;
      for (int i = 1; i < n; i++) {
        for (int j = 0; j <= split; j++) {
          if (j > 0 && s[i] == t[j - 1] && dp[i - 1][j - 1] >= 0)
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          if (split + dp[i - 1][j] < m && s[i] == t[split + dp[i - 1][j]] &&
              dp[i - 1][j] >= 0)
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
          dp[i][j] = max(dp[i - 1][j], dp[i][j]);
        }
      }
      if (dp[n - 1][split] >= m - split) chk = 1;
    }
    if (chk)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
