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 INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int N = 405;
bool check(string s1, string s2) {
  int j = 0;
  for (int i = 0; i < s1.size() && j < s2.size(); i++) {
    if (s1[i] == s2[j]) j++;
  }
  return j == s2.size();
}
int f[N];
int main() {
  int _;
  cin >> _;
  while (_--) {
    string s1, s2, s3;
    cin >> s1 >> s2;
    int n = s1.size(), m = s2.size();
    int ok = 0;
    if (check(s1, s2)) {
      puts("YES");
      continue;
    }
    for (int i = 0; i < m - 1; i++) {
      string t1 = s2.substr(0, i + 1);
      string t2 = s2.substr(i + 1);
      memset(f, -1, sizeof f);
      f[0] = 0;
      for (int j = 0; j < n; j++) {
        for (int k = t1.size(); k >= 0; k--) {
          if (f[k] != -1 && f[k] != t2.size() && s1[j] == t2[f[k]]) f[k]++;
          if (k >= 1 && s1[j] == t1[k - 1]) f[k] = max(f[k], f[k - 1]);
        }
      }
      if (f[t1.size()] == t2.size()) {
        ok = 1;
        break;
      }
    }
    puts(ok ? "YES" : "NO");
  }
  return 0;
}
