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;
using lol = long long int;
const lol mod1 = 1e9 + 7, mod2 = 998244353;
const lol inf = 1e12 + 8;
const double eps = 1e-12;
const int N = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int _ = 1;
  cin >> _;
  while (_--) {
    string s, t;
    cin >> s >> t;
    if (t.size() == 1) {
      cout << ((s.find(t) == string::npos) ? "NO" : "YES") << "\n";
      continue;
    }
    bool found = 0;
    for (int len = 1; len < t.size(); len++) {
      string q1 = t.substr(0, len), q2 = t.substr(len);
      int dp[s.size() + 1][q1.size() + 1];
      for (int i = 0; i <= s.size(); i++) {
        dp[i][0] = 0;
        for (int j = 1; j <= q1.size(); j++) {
          dp[i][j] = -1;
        }
      }
      for (int i = 1; i <= s.size(); i++) {
        for (int j = 0; j <= q1.size(); j++) {
          dp[i][j] = dp[i - 1][j] + (dp[i - 1][j] != q2.size() &&
                                     q2[dp[i - 1][j]] == s[i - 1]);
          if (j > 0) {
            if (q1[j - 1] == s[i - 1]) {
              dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
            }
          }
        }
      }
      if (dp[s.size()][q1.size()] == q2.size()) {
        found = 1;
        break;
      }
    }
    cout << ((found) ? "YES" : "NO") << "\n";
  }
  return 0;
}
