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;
long long f[405][405], n, t;
char s[405], o[405];
bool dp(long long h) {
  for (long long i = 0; i <= h; i++) fill(f[i] + i + 1, f[i] + h + 1, -99999);
  for (long long i = 1; i <= n; i++)
    for (long long j = min(i, h), u; j >= 0; j--) {
      u = h + f[i - 1][j] + 1;
      f[i][j] = f[i - 1][j] + (u > 0 && u <= t && s[i] == o[u]);
      if (j && s[i] == o[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
    }
  return f[n][h] == t - h;
}
bool can() {
  for (long long i = 0; i < t; i++)
    if (dp(i)) return 1;
  return 0;
}
int32_t main() {
  long long T;
  cin >> T;
  while (T--) {
    cin >> s + 1 >> o + 1;
    n = strlen(s + 1), t = strlen(o + 1);
    if (can())
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
