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;
using ll = long long;
const ll inf = 1e18;
const int N = 2 * 1e5 + 10;
ll en;
ll res;
ll a[N], h[6996];
int dp[404][404][404];
string s, t;
bool go(ll i, ll ii, ll iii) {
  if (iii == s.size()) return (i == en and ii == t.size());
  if (i == en and ii == t.size()) return dp[iii][i][ii] = 1;
  if (dp[iii][i][ii] + 1) return dp[iii][i][ii];
  bool fl = false;
  if (i < en and s[iii] == t[i]) fl |= go(i + 1, ii, iii + 1);
  if (ii < t.size() and s[iii] == t[ii]) fl |= go(i, ii + 1, iii + 1);
  fl |= go(i, ii, iii + 1);
  return dp[iii][i][ii] = fl;
}
void solve() {
  cin >> s >> t;
  for (int i = 0; i <= s.size(); ++i) {
    for (int j = 0; j <= t.size(); ++j) {
      for (int k = 0; k <= t.size(); ++k) {
        dp[i][j][k] = -1;
      }
    }
  }
  for (int i = 0; i <= t.size(); ++i) {
    en = i;
    if (go(0, i, 0)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
