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;
string s1, s2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    cin >> s1 >> s2;
    for (long long i = 0; i < s2.size(); ++i) {
      vector<long long> cur(s2.size(), -1);
      vector<long long> nxt(s2.size(), -1);
      cur[0] = i;
      for (long long j = 0; j < s1.size(); ++j) {
        for (long long k = 0; k <= i; ++k) {
          if (cur[k] == -1) continue;
          if (k == i && cur[k] == s2.size()) {
            cout << "YES\n";
            goto END;
          }
          nxt[k] = max(nxt[k], cur[k]);
          if (cur[k] < s2.size() && s1[j] == s2[cur[k]])
            nxt[k] = max(nxt[k], cur[k] + 1);
          if (k < i && s1[j] == s2[k]) nxt[k + 1] = max(nxt[k + 1], cur[k]);
        }
        cur = nxt;
      }
      if (cur[i] == s2.size()) {
        cout << "YES\n";
        goto END;
      }
    }
    cout << "NO\n";
  END:;
  }
}
