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;
int dp[401][401][401];
string s, t;
int f(int i, int j, int k) {
  if (i == s.size()) {
    if (k == t.size()) return j;
    return -2;
  }
  int &ret = dp[i][j][k];
  if (~ret) return ret;
  ret = f(i + 1, j, k);
  if (j < t.size() && s[i] == t[j]) ret = max(ret, f(i + 1, j + 1, k));
  if (k < t.size() && s[i] == t[k]) ret = max(ret, f(i + 1, j, k + 1));
  return ret;
}
void solve() {
  cin >> s >> t;
  for (int i = 0; i <= s.size(); i++)
    for (int j = 0; j <= t.size(); j++)
      for (int k = 0; k <= t.size(); k++) dp[i][j][k] = -1;
  for (int i = 0; i < s.size(); i++) {
    if (f(0, 0, i) >= i) {
      cout << "YES" << '\n';
      return;
    }
  }
  cout << "NO" << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
