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 long long int N = 100001;
const long long int M = 1;
struct point {
  long long int x, y;
  long long int i;
};
long long int fast_exp(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return fast_exp(x, n - 1) * x % 998244353;
  else {
    long long int tmp = fast_exp(x, n >> 1);
    return tmp * tmp % 998244353;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int _ = 1;
  cin >> _;
  while (_--) {
    long long int n, m, x = 0, y, ans = 0;
    string s, goal;
    cin >> s >> goal;
    n = s.size();
    m = goal.size();
    long long int nxt[n][26];
    for (int i = n - 1; i > -1; i--) {
      if (i != n - 1)
        for (int j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
      else
        for (int j = 0; j < 26; j++) nxt[i][j] = n;
      nxt[i][s[i] - 'a'] = i;
    }
    for (int i = 0; i < m; i++) {
      x = i + 1;
      y = m - x;
      long long int dp[x + 1][y + 1];
      for (int j = 0; j <= x; j++)
        for (int k = 0; k <= y; k++) dp[j][k] = n;
      dp[0][0] = -1;
      for (int j = 0; j <= x; j++) {
        for (int k = 0; k <= y; k++) {
          if (j > 0 && dp[j - 1][k] < n - 1)
            dp[j][k] = nxt[dp[j - 1][k] + 1][goal[j - 1] - 'a'];
          if (k > 0 && dp[j][k - 1] < n - 1)
            dp[j][k] =
                min(dp[j][k], nxt[dp[j][k - 1] + 1][goal[x + k - 1] - 'a']);
        }
      }
      if (dp[x][y] < n) ans = 1;
    }
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
