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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
string s, t;
long long nxt[401][26];
bool solve(string a, string b) {
  long long n = a.size(), m = b.size();
  long long dp[n + 1][m + 1];
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= m; j++) dp[i][j] = 1000000000000000000;
  }
  dp[0][0] = 0;
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= m; j++) {
      if (dp[i][j] > s.size()) continue;
      long long len = dp[i][j];
      if (i < n && nxt[len][a[i] - 'a'] < 1000000000000000000) {
        dp[i + 1][j] = min(dp[i + 1][j], nxt[len][a[i] - 'a'] + 1);
      }
      if (j < m && nxt[len][b[j] - 'a'] < 1000000000000000000) {
        dp[i][j + 1] = min(dp[i][j + 1], nxt[len][b[j] - 'a'] + 1);
      }
    }
  }
  if (dp[n][m] < 1000000000000000000)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    for (long long i = 0; i <= s.size(); i++) {
      for (long long j = 0; j < 26; j++) nxt[i][j] = 1000000000000000000;
    }
    for (long long i = s.size() - 1; i >= 0; i--) {
      for (long long j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
      nxt[i][s[i] - 'a'] = i;
    }
    bool flag = true;
    for (long long i = 0; i < t.size(); i++) {
      string a, b;
      for (long long j = 0; j < t.size(); j++) {
        if (j < i)
          a += t[j];
        else
          b += t[j];
      }
      if (solve(a, b)) {
        cout << "YES";
        flag = false;
        break;
      }
    }
    if (flag) cout << "NO";
    cout << "\n";
  }
  return 0;
}
