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 mem[405][405][405];
string s, t;
int n, m;
int go(int i, int j, int k) {
  if (j >= m && k >= n) return i;
  if (k >= n) return -10;
  if (mem[i][j][k] != -1) return mem[i][j][k];
  mem[i][j][k] = go(i, j, k + 1);
  if (i < m && t[i] == s[k])
    mem[i][j][k] = max(mem[i][j][k], go(i + 1, j, k + 1));
  if (j < m && t[j] == s[k])
    mem[i][j][k] = max(mem[i][j][k], go(i, j + 1, k + 1));
  return mem[i][j][k];
}
int main(int argc, const char *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    cin >> s >> t;
    n = s.size();
    m = t.size();
    for (int i = 0; i <= m; i++)
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= n; k++) mem[i][j][k] = -1;
    int ans = 0;
    for (int i = 0; i < m && (!ans); i++)
      if (go(0, i, 0) >= i) ans = 1;
    cout << (ans ? "YES" : "NO") << endl;
  }
  return 0;
}
