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;
namespace Greedy {
bool solve(const string& S, const string& T) {
  int j = int((T).size()) - 1;
  const int N = int((S).size());
  vector<int> used(N, 0);
  for (int op = 1; op <= 2; ++op) {
    for (int i = N - 1; i >= 0; --i) {
      if (used[i] > 0) continue;
      if (S[i] == T[j]) {
        used[i] = op;
        --j;
        if (j < 0) return true;
      }
    }
  }
  return false;
}
}  // namespace Greedy
namespace DP {
bool trace;
string S, T1, T2;
unsigned short dp_id;
unsigned short cache[402][402][402];
bool memo[402][402][402];
bool go(int n, int i, int j, int k) {
  if (T1[i] == '$' && T2[j] == '$') return 1;
  if (n >= int((S).size()) || k < 0) return 0;
  bool& res = memo[i][j][k];
  if (cache[i][j][k] != dp_id) {
    cache[i][j][k] = dp_id;
    res = 0;
    if (S[n] == T1[i]) {
      bool cur = go(n + 1, i + 1, j, k);
      if (cur) return res = true;
    }
    if (S[n] == T2[j]) {
      bool cur = go(n + 1, i, j + 1, k);
      if (cur) return res = true;
    }
    if (S[n] != T1[i] && S[n] != T2[j]) {
      bool cur = go(n + 1, i, j, k - 1);
      if (cur) return res = true;
    }
  }
  return res;
}
bool solve(int tc, const string& S, const string& T) {
  for (int i = 0, j = 0; i < int((S).size()); ++i) {
    if (S[i] == T[j]) {
      ++j;
      if (j >= int((T).size())) return true;
    }
  }
  DP::S = S;
  for (int m = 1; m < int((T).size()); ++m) {
    T1 = T.substr(0, m);
    T2 = T.substr(m);
    T1 += '$';
    T2 += '$';
    dp_id = tc * 404 + m;
    if (go(0, 0, 0, int((S).size()) - int((T).size()))) {
      return true;
    }
  }
  return false;
}
}  // namespace DP
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC;
  cin >> TC;
  for (int tc = (1), _b = (TC); tc <= _b; ++tc) {
    string S, T;
    cin >> S;
    cin >> T;
    bool res = DP::solve(tc, S, T);
    cout << (res ? "YES" : "NO") << '\n';
  }
  return 0;
}
