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;
bool ch(string& s, string& a, string& b, vector<vector<long long> >& prec) {
  long long dp[(long long)a.length() + 5][(long long)b.length() + 5];
  for (long long i = 0; i <= a.length(); i++) {
    for (long long j = 0; j <= b.length(); j++) {
      dp[i][j] = INT_MAX;
      if (!i and !j)
        dp[i][j] = -1;
      else {
        if (i and dp[i - 1][j] + 1 < s.length())
          dp[i][j] = min(dp[i][j], prec[dp[i - 1][j] + 1][a[i - 1] - 'a']);
        if (j and dp[i][j - 1] + 1 < s.length())
          dp[i][j] = min(dp[i][j], prec[dp[i][j - 1] + 1][b[j - 1] - 'a']);
      }
    }
  }
  if (dp[a.length()][b.length()] < s.length())
    return 1;
  else
    return 0;
}
void pre(string& s, vector<vector<long long> >& prec) {
  prec.assign(s.length() + 1, vector<long long>(30, INT_MAX));
  for (long long i = s.length() - 1; i >= 0; i--) {
    prec[i][s[i] - 'a'] = i;
    for (long long j = 0; j < 26; j++) {
      if (i + 1 < s.length()) prec[i][j] = min(prec[i][j], prec[i + 1][j]);
    }
  }
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    string s, p;
    vector<vector<long long> > prec;
    cin >> s >> p;
    pre(s, prec);
    bool poss = 0;
    for (long long i = 0; i < p.length(); i++) {
      string a = p.substr(0, i + 1);
      string b = p.substr(i + 1, p.length() - i - 1);
      poss |= ch(s, a, b, prec);
    }
    if (poss)
      cout << "YES";
    else
      cout << "NO";
    cout << endl;
  }
}
