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 n, m;
int dp[403][403][403];
string str, str1;
int solve(int i, int j, int k) {
  if (i == n) {
    if (k == m) {
      return (j);
    } else {
      return (-1);
    }
  }
  if (dp[i][j][k] != -2) {
    return (dp[i][j][k]);
  }
  int ans = -1;
  if (j < m && str1[j] == str[i]) {
    ans = max(ans, solve(i + 1, j + 1, k));
  }
  if (k < m && str1[k] == str[i]) {
    ans = max(ans, solve(i + 1, j, k + 1));
  }
  ans = max(ans, solve(i + 1, j, k));
  dp[i][j][k] = ans;
  return (ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int a, i, j, k, t;
  cin >> t;
  while (t--) {
    vector<int> arr(26, 0);
    cin >> str >> str1;
    n = str.size();
    m = str1.size();
    for (i = 0; i < n; i = i + 1) {
      arr[str[i] - 97]++;
    }
    for (i = 0; i < m; i = i + 1) {
      arr[str1[i] - 97]--;
    }
    a = 0;
    for (i = 0; i < 26; i = i + 1) {
      if (arr[i] < 0) {
        a = 1;
        break;
      }
    }
    if (a) {
      cout << "NO\n";
      continue;
    }
    for (i = 0; i <= n; i++) {
      for (j = 0; j <= m; j = j + 1) {
        for (k = 0; k <= m; k = k + 1) {
          dp[i][j][k] = -2;
        }
      }
    }
    a = 0;
    for (i = 0; i < m; i = i + 1) {
      if (solve(0, 0, i) >= i) {
        cout << "YES\n";
        a = 1;
        break;
      }
    }
    if (!a) {
      cout << "NO\n";
    }
  }
}
