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;
const int INF = 0x3f3f3f3f;
void solve() {
  string s, t;
  cin >> s >> t;
  int cnt = 0;
  for (int p = 1; p <= t.size(); p++) {
    string a = t.substr(0, p);
    string b = t.substr(p);
    vector<vector<int> > dp(s.size(), vector<int>(a.size() + 1, -INF));
    for (int i = 0; i < s.size(); i++) {
      for (int j = 0; j <= a.size(); j++) {
        int ans1 = i > 0 ? dp[i - 1][j] : (j > 0 ? -INF : 0);
        if (ans1 >= 0 && ans1 < b.size() && b[ans1] == s[i])
          dp[i][j] = max(dp[i][j], ans1 + 1);
        if (j > 0 && s[i] == a[j - 1])
          dp[i][j] =
              max(dp[i][j], i > 0 ? dp[i - 1][j - 1] : (j == 1 ? 0 : -INF));
        dp[i][j] = max(dp[i][j], ans1);
      }
    }
    if (dp[s.size() - 1][a.size()] == b.size()) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}
