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;
void _print() { cout << '\n'; }
template <typename T, typename... Ts>
void _print(const T& value, const Ts&... values) {
  cout << ' ' << value;
  _print(values...);
}
template <typename T, typename... Ts>
void print(const T& value, const Ts&... values) {
  cout << value;
  _print(values...);
}
string a, b, f1, f2;
vector<vector<int> > dp;
bool check() {
  dp.assign(a.size() + 1, vector<int>(f1.size() + 1, -1));
  dp[0][0] = 0;
  for (int len = (0); len <= (a.size() - 1); len++) {
    char c = a[len];
    for (int match1 = (0); match1 <= (f1.size()); match1++) {
      if (dp[len][match1] == -1) continue;
      if (match1 != f1.size() && c == f1[match1])
        dp[len + 1][match1 + 1] = max(dp[len + 1][match1 + 1], dp[len][match1]);
      if (dp[len][match1] != f2.size() && c == f2[dp[len][match1]])
        dp[len + 1][match1] = max(dp[len + 1][match1], dp[len][match1] + 1);
      dp[len + 1][match1] = max(dp[len + 1][match1], dp[len][match1]);
    }
  }
  if (dp[a.size()][f1.size()] == f2.size()) return true;
  return false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> a >> b;
    f1 = "";
    f2 = b;
    bool ok = false;
    for (int i = (1); i <= (b.size()); i++) {
      f1 += f2.front();
      f2.erase(f2.begin());
      if (check()) {
        ok = true;
        break;
      }
    }
    if (ok)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}
