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 T, nxt[410][26], f[410][410];
string s, t;
int main() {
  cin >> T;
  while (T--) {
    cin >> s >> t;
    memset(nxt, 0x3f, sizeof(nxt));
    for (int i = s.length() - 1; ~i; i--) {
      memcpy(nxt[i], nxt[i + 1], sizeof(nxt[i]));
      nxt[i][s[i] - 'a'] = i;
    }
    auto chk = [&](string a, string b) {
      memset(f, 0x3f, sizeof(f)), f[0][0] = 0;
      for (int i = 0; i <= a.length(); i++) {
        for (int j = 0; j <= b.length(); j++) {
          if (f[i][j] > s.length()) continue;
          if (i < a.length() && nxt[f[i][j]][a[i] - 'a'] < 1e9) {
            f[i + 1][j] = min(f[i + 1][j], nxt[f[i][j]][a[i] - 'a'] + 1);
          }
          if (j < b.length() && nxt[f[i][j]][b[j] - 'a'] < 1e9) {
            f[i][j + 1] = min(f[i][j + 1], nxt[f[i][j]][b[j] - 'a'] + 1);
          }
        }
      }
      return f[a.length()][b.length()] < 1e9;
    };
    bool flag = 0;
    for (int i = 0; i < t.length(); i++) {
      if (chk(t.substr(0, i), t.substr(i, t.length()))) {
        flag = 1;
        break;
      }
    }
    cout << (flag ? "YES" : "NO") << endl;
  }
  return 0;
}
