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 double pi = 3.141592653589793238;
int MOD = 1000000007;
const int N = 1e6 + 10;
int n, m;
string s, t;
int dp[401][401][401];
int K;
int rec(int i, int j, int itr) {
  if (i == K && j == m) return 1;
  if (itr == n) return 0;
  int& ans = dp[i][j][itr];
  if (ans != -1) return ans;
  ans = 0;
  if (i < K && s[itr] == t[i]) ans |= rec(i + 1, j, itr + 1);
  if (j < m && s[itr] == t[j]) ans |= rec(i, j + 1, itr + 1);
  ans |= rec(i, j, itr + 1);
  return ans;
}
void solve() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  bool ok = 0;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++)
      for (int itr = 0; itr <= n; itr++) dp[i][j][itr] = -1;
  for (int k = 0; k <= m; k++) {
    K = k;
    ok |= rec(0, k, 0);
    if (ok) break;
  }
  cout << (ok ? "YES" : "NO") << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
