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;
#pragma GCC diagnostic ignored "-Wmissing-declarations"
bool check(const string& s, const string& t1, const string& t2) {
  vector<int> dp(t1.size() + 1, -1);
  dp[0] = 0;
  for (char c : s) {
    auto ndp = dp;
    for (int i = 0; i <= t1.size(); ++i)
      if (dp[i] >= 0 && dp[i] < t2.size() && t2[dp[i]] == c)
        ndp[i] = max(ndp[i], dp[i] + 1);
    for (int i = 0; i < t1.size(); ++i)
      if (dp[i] >= 0 && t1[i] == c) ndp[i + 1] = max(ndp[i + 1], dp[i]);
    dp.swap(ndp);
  }
  return dp.back() == t2.size();
}
void solve() {
  string s, p;
  cin >> s >> p;
  for (size_t i = 0; i < p.size(); ++i)
    if (check(s, p.substr(0, i), p.substr(i, p.size() - i))) {
      cout << "YES\n";
      return;
    }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
