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 N = 405;
int dp[N][N][N], n, flg, m;
string s, t;
int func(int i, int j, int k) {
  if (i == n && k == 0) return 0;
  if (i == n) return -1e9;
  if (dp[i][j][k] != -1) return dp[i][j][k];
  int &ans = dp[i][j][k];
  ans = 0;
  if (i < n) ans = func(i + 1, j, k);
  if (i < n && s[i] == t[j]) ans = max(ans, 1 + func(i + 1, j + 1, k));
  if (i < n && k > 0 && s[i] == t[m - k]) ans = max(ans, func(i + 1, j, k - 1));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tt;
  cin >> tt;
  while (tt--) {
    cin >> s >> t;
    n = s.size(), m = t.size(), flg = 0;
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++)
        for (int k = 0; k <= m; k++) dp[i][j][k] = -1;
    for (int i = 0; i <= m; i++) {
      if (func(0, 0, i) >= m - i) {
        flg = 1;
        break;
      }
    }
    cout << (flg ? "YES\n" : "NO\n");
  }
}
