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;
const int kMaxn = 405;
int f[kMaxn][kMaxn], n, m;
bool check(int x) {
  memset(f, 0, sizeof f);
  f[0][0] = x - 1;
  for (int i = 1; i <= n; ++i) {
    f[i][0] = f[i - 1][0];
    if (s[i] == t[f[i - 1][0] + 1]) f[i][0] = f[i - 1][0] + 1;
    for (int j = 1; j < x; ++j) {
      f[i][j] = f[i - 1][j];
      if (t[j] == s[i]) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (t[f[i - 1][j] + 1] == s[i]) f[i][j] = max(f[i][j], f[i - 1][j] + 1);
    }
  }
  return f[n][x - 1] == m;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    n = s.size();
    m = t.size();
    s = ' ' + s;
    t = ' ' + t;
    for (int i = 1; i <= m + 1; ++i)
      if (check(i)) {
        cout << "YES\n";
        goto s;
      }
    cout << "NO\n";
  s:;
  }
  return 0;
}
