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>
const long long INF = 1e9 + 7, mod = 998244353, MAXN = 1e5, SQ = 450,
                pp = 1e5 + 3;
const long double EPS = 1e-9, PI = acos(-1);
using namespace std;
namespace in_out {
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto i : a) out << i << " ";
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
}  // namespace in_out
using namespace in_out;
signed main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    string second, t;
    cin >> second >> t;
    vector<vector<int>> next(second.size() + 1, vector<int>(26, INF));
    for (int i = (int)second.size() - 1; i >= 0; i--) {
      next[i] = next[i + 1];
      next[i][second[i] - 'a'] = i;
    }
    bool first = 0;
    for (int k = 0; k < (int)t.size(); k++) {
      string a = t.substr(0, k);
      string b = t.substr(k, t.size());
      vector<vector<int>> dp(a.size() + 1, vector<int>(b.size() + 1, INF));
      dp[0][0] = 0;
      for (int i = 0; i <= (int)a.size(); i++)
        for (int j = 0; j <= (int)b.size(); j++) {
          if (dp[i][j] == INF) continue;
          if (i < (int)a.size())
            dp[i + 1][j] = min(dp[i + 1][j], next[dp[i][j]][a[i] - 'a'] + 1);
          if (j < (int)b.size())
            dp[i][j + 1] = min(dp[i][j + 1], next[dp[i][j]][b[j] - 'a'] + 1);
        }
      first |= dp[a.size()][b.size()] < INF;
    }
    if (first)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
