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 mod = 1e9 + 7;
const int N = 1e5 + 6;
const long long mx = 1e9;
int dp[410][410];
string s, t;
void solve() {
  int n, m;
  cin >> s >> t;
  n = s.length();
  m = t.length();
  for (int k = 0; k <= m - 1; k++) {
    int y = m - k;
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 0; i <= n - 1; i++)
      for (int j = 0; j <= k; j++) {
        if (dp[i][j] == -1) continue;
        if (j < k && s[i] == t[j])
          dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + (t[dp[i][j] + k] == s[i]));
      }
    if (dp[n][k] == y) {
      puts("YES");
      return;
    }
  }
  puts("NO");
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) solve();
  return 0;
}
