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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  }
  return os;
}
void solve() {
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int ts = t.size();
  vector<vector<int>> nxt(n + 1, vector<int>(26, -1));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; (j) < (int)(26); (j)++) {
      nxt[i][j] = nxt[i + 1][j];
    }
    nxt[i][s[i] - 'a'] = i + 1;
  }
  bool ok = false;
  for (int m = 0; (m) < (int)(ts + 1); (m)++) {
    string t0 = t.substr(0, m);
    string t1 = t.substr(m, n - m);
    int p = t0.size();
    int q = t1.size();
    vector<vector<int>> dp(p + 1, vector<int>(q + 1, 1 << 30));
    dp[0][0] = 0;
    for (int i = 0; (i) < (int)(p + 1); (i)++) {
      for (int j = 0; (j) < (int)(q + 1); (j)++) {
        if (dp[i][j] != 1 << 30) {
          if (i < p && nxt[dp[i][j]][t0[i] - 'a'] != -1) {
            dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j]][t0[i] - 'a']);
          }
          if (j < q && nxt[dp[i][j]][t1[j] - 'a'] != -1) {
            dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j]][t1[j] - 'a']);
          }
        }
      }
    }
    if (dp[p][q] != 1 << 30) {
      ok = true;
      break;
    }
  }
  cout << (ok ? "YES\n" : "NO\n");
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
