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 c = 402;
int regi[c][c], uj[c][c], w, n, m;
string a, b;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> w;
  while (w--) {
    cin >> a >> b;
    n = a.size(), m = b.size();
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        regi[i][j] = -1;
        uj[i][j] = -1;
      }
      regi[i][i] = 0;
    }
    for (int i = 0; i < n; i++) {
      for (int kezd = 1; kezd <= m; kezd++) {
        for (int veg = kezd; veg <= m; veg++) {
          int s = regi[kezd][veg];
          if (s != -1) {
            uj[kezd][veg] = max(uj[kezd][veg], s);
            if (s < m && a[i] == b[s]) {
              uj[kezd][veg] = max(uj[kezd][veg], s + 1);
            }
            if (veg < m && a[i] == b[veg]) {
              uj[kezd][veg + 1] = max(uj[kezd][veg + 1], s);
            }
          }
        }
      }
      for (int kezd = 1; kezd <= m; kezd++) {
        for (int veg = 1; veg <= m; veg++) {
          regi[kezd][veg] = uj[kezd][veg];
          uj[kezd][veg] = -1;
        }
      }
    }
    bool jo = 0;
    for (int kezd = 1; kezd <= m; kezd++) {
      if (regi[kezd][m] >= kezd) {
        jo = 1;
      }
    }
    cout << (jo ? "YES" : "NO") << "\n";
  }
  return 0;
}
