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;
template <typename T>
istream& operator>>(istream& is, vector<T>& a) {
  for (T& ai : a) is >> ai;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
  os << "[ ";
  for (const T& ai : a) os << ai << " ";
  return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
  return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
bool f() {
  string s, t;
  cin >> s >> t;
  int n = s.size();
  vector<vector<int> > p(n);
  p.push_back(vector<int>(26, n));
  for (int i = n - 1; i >= 0; --i) {
    p[i] = p[i + 1];
    p[i][s[i] - 'a'] = i;
  }
  int m = t.size();
  if (m == 1) return p[0][t[0] - 'a'] < n;
  for (int m1 = 1; m1 < m; ++m1) {
    int m2 = m - m1;
    vector<vector<int> > dp(m1 + 1, vector<int>(m2 + 1, n + 1));
    dp[0][0] = 0;
    for (int i1 = 0; i1 <= m1; ++i1) {
      for (int i2 = 0; i2 <= m2; ++i2) {
        if (dp[i1][i2] >= n) continue;
        if (i1 < m1)
          dp[i1 + 1][i2] = min(dp[i1 + 1][i2], p[dp[i1][i2]][t[i1] - 'a'] + 1);
        if (i2 < m2)
          dp[i1][i2 + 1] =
              min(dp[i1][i2 + 1], p[dp[i1][i2]][t[m1 + i2] - 'a'] + 1);
      }
    }
    if (dp[m1][m2] <= n) return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) cout << (f() ? "YES\n" : "NO\n");
  return 0;
}
