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 T;
string s, t;
long long dp[409][409];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    cin >> s >> t;
    long long ok = 0;
    for (long long i = 0; i < (long long)(t).size(); ++i) {
      string a = t.substr(0, i + 1);
      string b = t.substr(i + 1, (long long)(t).size() - i - 1);
      for (long long j = 0; j <= (long long)(s).size(); ++j)
        for (long long k = 0; k <= (long long)(a).size(); ++k) dp[j][k] = -1e9;
      dp[0][0] = 0;
      for (long long j = 1; j <= (long long)(s).size(); ++j)
        for (long long k = 0; k <= (long long)(a).size() && k <= j; ++k) {
          dp[j][k] = dp[j - 1][k];
          if (k != 0 && a[k - 1] == s[j - 1])
            dp[j][k] = max(dp[j][k], dp[j - 1][k - 1]);
          if ((long long)(b).size() >= 1 && dp[j - 1][k] >= 0 &&
              b[dp[j - 1][k]] == s[j - 1])
            dp[j][k] = max(dp[j][k], dp[j - 1][k] + 1);
        }
      if (dp[(long long)(s).size()][(long long)(a).size()] ==
          (long long)(b).size())
        ok = 1;
    }
    if (ok)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}
