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 long long inf = 1e15;
const long long mod = 1e9 + 7;
int sz;
short int dp[409][409][409];
string s, t;
short int bt(int id, int id1, int id2) {
  if (id == s.size()) return (id1 == sz && id2 == t.size());
  short int &ret = dp[id][id1][id2];
  if (ret != -1) return ret;
  ret = bt(id + 1, id1, id2);
  if (id1 < sz && t[id1] == s[id]) {
    ret = max(ret, bt(id + 1, id1 + 1, id2));
  }
  if (id2 < t.size() && t[id2] == s[id]) {
    ret = max(ret, bt(id + 1, id1, id2 + 1));
  }
  return ret;
}
void solve() {
  cin >> s >> t;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < t.size(); i++) {
    sz = i;
    if (bt(0, 0, i) == 1) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int main() {
  int t;
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> t;
  while (t--) solve();
}
