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 ll = long long;
void assign_max(int &a, int b) { a = max(a, b); }
bool check(const string &s, const string &a, const string &b) {
  vector<vector<int>> d(s.size() + 1, vector<int>(a.size() + 1, -1));
  d[0][0] = 0;
  for (int i = 0; i < (int)s.size(); ++i) {
    for (int ia = 0; ia < (int)a.size(); ++ia) {
      int t = d[i][ia];
      if (t == -1) {
        continue;
      }
      assign_max(d[i + 1][ia], t + (t == (int)b.size() ? 0 : s[i] == b[t]));
      if (s[i] == a[ia]) {
        assign_max(d[i + 1][ia + 1], t);
      }
    }
    int t = d[i][a.size()];
    assign_max(d[i + 1][a.size()], t + (t == (int)b.size() ? 0 : s[i] == b[t]));
  }
  return d[s.size()][a.size()] == (int)b.size();
}
void solve() {
  string s, t;
  cin >> s >> t;
  for (int i = 0; i <= (int)t.size(); ++i) {
    if (check(s, t.substr(0, i), t.substr(i))) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
