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 double PI = 3.1415926535898;
const double E = 2.718281828459;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int f[410];
int main(int argc, const char *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    string s, t;
    cin >> s >> t;
    int ls = s.length();
    int lt = t.length();
    int fl = 0;
    for (int p = 0; p < lt && (!fl); p++) {
      memset(f, -1, sizeof(f));
      f[p] = 0;
      for (int i = 0; i < ls; i++) {
        for (int j = lt; j >= p; j--) {
          if (f[j] < 0) continue;
          if (j < lt && s[i] == t[j]) f[j + 1] = max(f[j + 1], f[j]);
          if (f[j] < p && s[i] == t[f[j]]) f[j]++;
        }
      }
      if (f[lt] >= p) fl = 1;
    }
    if (fl)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
