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;
using ll = long long;
const int N = 400;
int dp[N + 1][N + 1];
bool solve(string str, string t1, string t2) {
  int n = str.size();
  int m = t1.size();
  for (int j = 1; j <= m; ++j) dp[0][j] = -1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      dp[i][j] = -1;
      if (dp[i - 1][j] != -1) {
        dp[i][j] = dp[i - 1][j];
        if (dp[i][j] < t2.size() && str[i - 1] == t2[dp[i][j]]) ++dp[i][j];
      }
      if (j > 0 && str[i - 1] == t1[j - 1]) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  return dp[n][m] >= (int)t2.size();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  for (int ti = 0; ti < tc; ++ti) {
    string src, tar;
    cin >> src >> tar;
    bool works = false;
    for (int a = 1; a <= tar.size(); ++a) {
      string le, ri;
      for (int i = 0; i < a; ++i) le.push_back(tar[i]);
      for (int i = a; i < tar.size(); ++i) ri.push_back(tar[i]);
      if (solve(src, le, ri)) works = true;
    }
    if (works)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
