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 MOD = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    if (s == t) {
      cout << "YES\n";
      continue;
    }
    int n = s.length();
    int m = t.length();
    int j = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == t[j]) {
        ++j;
        if (j == m) break;
      }
    }
    if (j == m) {
      cout << "YES\n";
      continue;
    }
    int flag = 1;
    for (int len = 1; len < m; len++) {
      string a = t.substr(0, len);
      string b = t.substr(len);
      int dp[2][len + 1];
      for (int j = 1; j <= len; j++) dp[0][j] = -1;
      dp[0][0] = 0;
      int r = 0;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j <= len; j++) {
          dp[r ^ 1][j] = dp[r][j];
          if (dp[r][j] >= 0 && dp[r][j] < b.length() && b[dp[r][j]] == s[i])
            ++dp[r ^ 1][j];
          if (dp[r][j - 1] >= 0 && j > 0 && a[j - 1] == s[i])
            dp[r ^ 1][j] = max(dp[r ^ 1][j], dp[r][j - 1]);
        }
        r ^= 1;
      }
      if (dp[r][len] == b.length()) {
        cout << "YES\n";
        flag = 0;
        break;
      }
    }
    if (flag) cout << "NO\n";
  }
}
