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 Max = 400 + 40 + 4;
int dp[Max][Max];
int nx[Max][26];
;
void solve() {
  string s, t;
  cin >> s >> t;
  memset(nx, 63, sizeof nx);
  int n = s.size(), m = t.size();
  for (int i = 0; i < n; i++)
    for (int j = n - 1; j >= i; j--) nx[i][s[j] - 'a'] = j + 1;
  for (int mid = 0; mid < m; mid++) {
    memset(dp, 63, sizeof dp);
    dp[0][mid] = 0;
    for (int sum = 0; sum <= m; sum++)
      for (int i = 0; i <= sum; i++) {
        int j = sum - i + mid;
        if (i > mid || j > m || dp[i][j] > n) continue;
        if (i < mid) dp[i + 1][j] = min(dp[i + 1][j], nx[dp[i][j]][t[i] - 'a']);
        if (j < m) dp[i][j + 1] = min(dp[i][j + 1], nx[dp[i][j]][t[j] - 'a']);
      }
    if (dp[mid][m] <= n) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
