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;
int T;
int dp[444];
int main() {
  cin >> T;
  for (; T--;) {
    string s, t;
    cin >> s >> t;
    bool ok = false;
    for (int k = 0; k <= t.size(); k++) {
      string t1 = t.substr(0, k), t2 = t.substr(k);
      for (int i = 0; i <= t1.size(); i++) dp[i] = -1;
      dp[0] = 0;
      for (int si = 0; si < s.size(); si++) {
        for (int i = min(si, (int)t1.size()); i >= 0; i--) {
          if (dp[i] < 0) continue;
          int j = dp[i];
          if (i < t1.size() && s[si] == t1[i])
            dp[i + 1] = max(dp[i + 1], dp[i]);
          if (j < t2.size() && s[si] == t2[j]) dp[i]++;
        }
      }
      if (dp[t1.size()] == t2.size()) {
        ok = true;
        break;
      }
    }
    cout << (ok ? "YES" : "NO") << endl;
  }
}
