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;
namespace DP {
const int INF = 1000000;
bool solve(int tc, string S, const string& T) {
  for (int i = 0, j = 0; i < int((S).size()); ++i) {
    if (S[i] == T[j]) {
      ++j;
      if (j >= int((T).size())) return true;
    }
  }
  S = "#" + S;
  vector<vector<int> > nxt(int((S).size()) + 1, vector<int>(26, INF));
  vector<int> last(26, INF);
  for (int n = int((S).size()) - 1; n >= 0; --n) {
    nxt[n] = last;
    int k = S[n] - 'a';
    last[k] = n;
  }
  for (int m = 1; m < int((T).size()); ++m) {
    string T1 = "#" + T.substr(0, m);
    string T2 = "#" + T.substr(m);
    vector<vector<int> > DP(int((T1).size()),
                            vector<int>(int((T2).size()), INF));
    DP[0][0] = 0;
    for (int i = 0, _n = (int((T1).size())); i < _n; ++i) {
      for (int j = 0, _n = (int((T2).size())); j < _n; ++j) {
        int n = DP[i][j];
        if (n >= int((S).size())) continue;
        if (i + 1 < int((T1).size())) {
          int k1 = T1[i + 1] - 'a';
          int n1 = nxt[n][k1];
          DP[i + 1][j] = min(DP[i + 1][j], n1);
        }
        if (j + 1 < int((T2).size())) {
          int k2 = T2[j + 1] - 'a';
          int n2 = nxt[n][k2];
          DP[i][j + 1] = min(DP[i][j + 1], n2);
        }
      }
    }
    if (DP[int((T1).size()) - 1][int((T2).size()) - 1] <= int((S).size())) {
      return true;
    }
  }
  return false;
}
}  // namespace DP
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC;
  cin >> TC;
  for (int tc = (1), _b = (TC); tc <= _b; ++tc) {
    string S, T;
    cin >> S;
    cin >> T;
    bool res = DP::solve(tc, S, T);
    cout << (res ? "YES" : "NO") << '\n';
  }
  return 0;
}
