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;
int d[402][402];
int c[402][26];
string A, B;
int n, m;
int go(string a, string b) {
  d[0][0] = -1;
  for (int i = 0; i <= a.size(); i++) {
    for (int j = 0; j <= b.size(); j++) {
      if (i == 0 && j == 0) continue;
      d[i][j] = n;
      if (i != 0) {
        d[i][j] = min(d[i][j], c[d[i - 1][j] + 1][(a[i - 1] - 'a')]);
      }
      if (j != 0) {
        d[i][j] = min(d[i][j], c[d[i][j - 1] + 1][(b[j - 1] - 'a')]);
      }
    }
  }
  if (d[a.size()][b.size()] < n) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> A >> B;
    bool f = false;
    n = A.size();
    for (int i = 0; i < 26; i++) {
      c[n][i] = n;
      c[n + 1][i] = n;
    }
    for (int i = n - 1; i >= 0; i--) {
      memcpy(c[i], c[i + 1], sizeof(c[0]));
      c[i][A[i] - 'a'] = i;
    }
    for (int i = 0; i < B.size(); i++) {
      if (go(B.substr(0, i), B.substr(i))) {
        f = true;
        cout << "YES" << '\n';
        break;
      }
    }
    if (!f) {
      cout << "NO" << '\n';
    }
  }
}
