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 int mod = 1e9 + 7;
int T, dp[405][405][405], k;
string s, t;
int f(int x, int y, int z, int og) {
  if (z == s.size() && y == t.size() && x == og) return k;
  if (z == s.size()) return k - 1;
  if (dp[x][y][z] >= k - 1) return dp[x][y][z];
  int ans = 0;
  if (x != og && t[x] == s[z]) ans = max(ans, f(x + 1, y, z + 1, og));
  if (y != t.size() && t[y] == s[z]) ans = max(ans, f(x, y + 1, z + 1, og));
  ans = max(ans, f(x, y, z + 1, og));
  return dp[x][y][z] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> s >> t;
    bool is = false;
    k += 2;
    for (int i = 0; i < s.size(); i++) {
      int x = f(0, i, 0, i);
      if (x == k) is = 1;
    }
    if (is)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}
