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;
int T;
string s, t;
int DP[405][405];
int main() {
  cin >> T;
  while (T--) {
    cin >> s >> t;
    int possible = false;
    for (int mid = 0; !possible && mid <= t.length(); mid++) {
      int lens = s.length();
      int lena = mid - 0;
      int lenb = t.length() - mid;
      string a = t.substr(0, lena);
      string b = t.substr(mid, lenb);
      if (false) {
        cout << "a = " << a << endl;
        cout << "b = " << b << endl;
      }
      for (int i = 0; i <= lena; i++) {
        for (int j = 0; j <= lenb; j++) {
          DP[i][j] = -1;
        }
      }
      DP[0][0] = s.length();
      for (int i = 1; i <= lena; i++) {
        int remain = DP[i - 1][0];
        while (remain > 0 && s[lens - remain] != a[i - 1]) remain -= 1;
        if (remain <= 0) break;
        DP[i][0] = remain - 1;
      }
      for (int j = 1; j <= lenb; j++) {
        int remain = DP[0][j - 1];
        while (remain > 0 && s[lens - remain] != b[j - 1]) remain -= 1;
        if (remain <= 0) break;
        DP[0][j] = remain - 1;
      }
      for (int i = 1; i <= lena; i++) {
        for (int j = 1; j <= lenb; j++) {
          int remain;
          remain = DP[i - 1][j];
          while (remain > 0 && s[lens - remain] != a[i - 1]) remain -= 1;
          DP[i][j] = max(DP[i][j], remain - 1);
          remain = DP[i][j - 1];
          while (remain > 0 && s[lens - remain] != b[j - 1]) remain -= 1;
          DP[i][j] = max(DP[i][j], remain - 1);
        }
      }
      if (false) {
        cout << "DP cache:" << endl;
        cout << "A/B\t";
        for (int i = 0; i <= lenb; i++) cout << i << '\t';
        cout << endl;
        for (int i = 0; i <= lena; i++) {
          cout << i << '\t';
          for (int j = 0; j <= lenb; j++) {
            cout << DP[i][j] << '\t';
          }
          cout << endl;
        }
      }
      if (DP[lena][lenb] >= 0) {
        possible = true;
      }
    }
    if (possible) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}
