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;
struct State {
  int x = 0, y = 0;
  State() {}
  State(int x, int y) : x(x), y(y) {}
};
bool check(string &s, int start, string &k) {
  vector<State> dp;
  dp.push_back(State(0, start));
  for (int i = 0; i < (int)s.size(); ++i) {
    vector<State> new_dp;
    for (State curr : dp) {
      if (curr.x != start && s[i] == k[curr.x])
        new_dp.push_back(State(curr.x + 1, curr.y));
      if (curr.y != (int)s.size() && s[i] == k[curr.y])
        new_dp.push_back(State(curr.x, curr.y + 1));
      new_dp.push_back(curr);
    }
    sort(new_dp.begin(), new_dp.end(), [](const State &A, const State &B) {
      if (A.x != B.x) return A.x < B.x;
      return A.y > B.y;
    });
    dp.clear();
    for (auto it : new_dp)
      if (dp.empty() || dp.back().x != it.x) {
        dp.push_back(it);
      }
    for (auto it : dp)
      if (it.x == start && it.y == (int)k.size()) return true;
  }
  return false;
}
void solve() {
  string s, k;
  cin >> s >> k;
  for (int i = 1; i < (int)k.size() + 1; ++i)
    if (check(s, i, k)) {
      cout << "YES\n";
      return;
    }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
