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;
using Int = long long;
int dp[500];
bool solve(const string &s, const string &t) {
  for (int n = 1; n <= t.size(); n++) {
    int m = t.size() - n;
    for (int i = 0; i < 500; i++) dp[i] = -1;
    dp[0] = 0;
    for (int i = 0; i < s.size(); i++) {
      for (int j = i; j >= 0; j--) {
        if (dp[j] == -1) continue;
        if (j + 1 <= n && s[i] == t[j]) {
          dp[j + 1] = max(dp[j + 1], dp[j]);
        }
        if (dp[j] + 1 <= m && s[i] == t[n + dp[j]]) {
          dp[j] += 1;
        }
      }
    }
    if (dp[n] == m) return true;
  }
  return false;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    string s, t;
    cin >> s >> t;
    cout << (solve(s, t) ? "YES\n" : "NO\n");
  }
  return 0;
}
