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 ML = 4E2;
string s, t;
int cur;
short memo[ML + 1][ML + 1][ML + 5];
bool solve(int i, int a, int b) {
  short& ans = memo[i][a][b];
  if (ans == -1) {
    if (a == cur && b == t.length()) {
      ans = true;
    } else {
      ans = false;
      if (i < s.length()) {
        if (a < cur && s[i] == t[a]) {
          ans |= solve(i + 1, a + 1, b);
        }
        if (b < t.length() && s[i] == t[b]) {
          ans |= solve(i + 1, a, b + 1);
        }
        ans |= solve(i + 1, a, b);
      }
    }
  }
  return ans;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int c;
  cin >> c;
  while (c--) {
    cin >> s >> t;
    bool possible = false;
    if (t.length() <= s.length()) {
      memset(memo, -1, sizeof(memo));
      for (int i = 0; !possible && i < t.length(); i++) {
        cur = i;
        possible |= solve(0, 0, cur);
      }
    }
    if (possible) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
