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 long long MAXN = 200 * 1000 + 321;
string s, t;
long long n, m;
inline bool Dipi(long long l) {
  long long dp[m + 1][l + 1];
  for (long long i = 0; i <= m; i++)
    for (long long j = 0; j <= l; j++) dp[i][j] = -1;
  dp[0][0] = 0;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 0; j <= l; j++) {
      dp[i][j] = dp[i - 1][j] + ((dp[i - 1][j] != -1 && dp[i - 1][j] != n - l)
                                     ? (t[l + dp[i - 1][j]] == s[i - 1])
                                     : false);
      if (j && s[i - 1] == t[j - 1]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  if (dp[m][l] == n - l)
    return true;
  else
    return false;
}
void solve() {
  cin >> s >> t;
  n = t.size();
  m = s.size();
  for (long long i = 0; i <= n; i++) {
    if (Dipi(i)) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
}
