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 int inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
void upd(int &x, int y) {
  if (x < y) x = y;
}
int T, n, m, f[405][405];
string s, t;
void solve() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  s = ' ' + s;
  t = ' ' + t;
  for (int(i) = (0); (i) <= (m); (i)++) {
    int u = m - i;
    for (int(j) = (0); (j) <= (n); (j)++)
      for (int(k) = (0); (k) <= (i); (k)++) f[j][k] = -1;
    f[0][0] = 0;
    for (int(j) = (0); (j) <= (n); (j)++)
      for (int(k) = (0); (k) <= (min(j, i)); (k)++) {
        if (f[j][k] == -1) continue;
        if (k < i && s[j + 1] == t[k + 1]) {
          upd(f[j + 1][k + 1], f[j][k]);
        }
        int cur = f[j][k];
        if (cur < u && t[i + cur + 1] == s[j + 1])
          upd(f[j + 1][k], cur + 1);
        else
          upd(f[j + 1][k], cur);
      }
    if (f[n][i] == u) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) solve();
  return 0;
}
