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;
string s, t;
int dp[401][401][401];
int rec(int i, int j, int k) {
  if (s.length() == i) {
    if (t.length() == k) return j;
    return -1e9;
  }
  if (dp[i][j][k] != -1) return dp[i][j][k];
  int ans = rec(i + 1, j, k);
  if (j < t.length() && s[i] == t[j]) ans = max(ans, rec(i + 1, j + 1, k));
  if (k < t.length() && s[i] == t[k]) ans = max(ans, rec(i + 1, j, k + 1));
  return dp[i][j][k] = ans;
}
void solve() {
  cin >> s >> t;
  int n = s.length() + 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < t.length() + 1; j++)
      for (int k = 0; k < t.length() + 1; k++) dp[i][j][k] = -1;
  for (int i = 0; i < t.length() + 1; i++) {
    if (rec(0, 0, i) >= i) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}
