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 N = 411;
int n, m, nxt[N][26], dp[N][N];
string s, t;
void minimize(int &x, const int &y) { x = min(x, y); }
void solve() {
  cin >> s >> t;
  n = (int)s.length(), m = (int)t.length();
  s = ' ' + s, t = ' ' + t;
  for (int i = 0; i < 26; ++i) nxt[n][i] = nxt[n + 1][i] = n + 1;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 0; j < 26; ++j) nxt[i][j] = nxt[i + 1][j];
    int c = s[i + 1] - 'a';
    nxt[i][c] = i + 1;
  }
  for (int i = 0; i < m; ++i) {
    memset(dp, 0x3f, sizeof(dp));
    dp[0][i] = 0;
    for (int j = 0; j <= i; ++j) {
      for (int k = i; k <= m; ++k) {
        if (dp[j][k] > n) continue;
        if (j < i) {
          int c = t[j + 1] - 'a';
          minimize(dp[j + 1][k], nxt[dp[j][k]][c]);
        }
        if (k < m) {
          int c = t[k + 1] - 'a';
          minimize(dp[j][k + 1], nxt[dp[j][k]][c]);
        }
      }
    }
    if (dp[i][m] <= n) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int nTest;
  cin >> nTest;
  while (nTest--) {
    solve();
  }
  return 0;
}
