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 ALP = 26;
const int INF = 1e5 + 239;
void solve() {
  string s, t;
  cin >> s >> t;
  vector<vector<int>> nx(ALP, vector<int>(s.size() + 1));
  for (int c = 0; c < ALP; c++) {
    nx[c][(int)s.size()] = INF;
    for (int i = (int)s.size() - 1; i >= 0; i--) {
      if (s[i] == (c + 'a'))
        nx[c][i] = i;
      else
        nx[c][i] = nx[c][i + 1];
    }
  }
  auto apply = [&](string x, string y) {
    vector<vector<int>> dp((int)x.size() + 1, vector<int>(y.size() + 1, INF));
    dp[0][0] = 0;
    for (int i = 0; i <= (int)x.size(); i++) {
      for (int j = 0; j <= (int)y.size(); j++) {
        if (dp[i][j] > (int)s.size()) continue;
        if (i + 1 <= (int)x.size()) {
          int to = nx[x[i] - 'a'][dp[i][j]];
          dp[i + 1][j] = min(dp[i + 1][j], to + 1);
        }
        if (j + 1 <= (int)y.size()) {
          int to = nx[y[j] - 'a'][dp[i][j]];
          dp[i][j + 1] = min(dp[i][j + 1], to + 1);
        }
      }
    }
    return dp[(int)x.size()][(int)y.size()] <= (int)s.size();
  };
  for (int lp = 0; lp < (int)t.size(); lp++) {
    string a = t.substr(0, lp);
    string b = t.substr(lp, (int)t.size() - lp);
    if (apply(a, b)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) solve();
}
