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;
long long n, m;
int dp[402][402];
int nxt[402][30];
int pos[30];
int main() {
  int t;
  cin >> t;
  while (t--) {
    string a;
    string b;
    cin >> a >> b;
    n = a.length();
    m = b.length();
    for (int i = 0; i < 26; i++) pos[i] = n + 1;
    for (int i = n + 1; i >= 0; i--) {
      for (int j = 0; j < 26; j++) {
        nxt[i][j] = pos[j];
      }
      if (i && i <= n) pos[a[i - 1] - 'a'] = i;
    }
    int is = 0;
    for (int x = 0; x <= m; x++) {
      for (int i = 0; i <= m; i++)
        for (int j = 0; j <= m; j++) dp[i][j] = n + 1;
      dp[0][0] = 0;
      for (int i = 0; i <= x; i++)
        for (int j = 0; j <= m - x; j++) {
          if (i < m)
            dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j]][b[i] - 'a']);
          if (j + x < m)
            dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j]][b[j + x] - 'a']);
        }
      if (dp[x][m - x] <= n) is = 1;
    }
    cout << (is ? "YES" : "NO") << endl;
  }
}
