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;
const long long N = 400;
long long dp[N + 1][N + 1];
int main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    bool possible = false;
    string s, t;
    cin >> s >> t;
    for (long long l = 0; l <= (long long)t.length() - 1; l++) {
      string p1 = "", p2 = "";
      for (long long i = 0; i <= l - 1; i++) {
        p1 += t[i];
      }
      for (long long i = l; i <= (long long)t.length() - 1; i++) {
        p2 += t[i];
      }
      dp[1][0] = (p2[0] == s[0]) ? 1 : 0;
      if (1 <= (long long)p1.length()) {
        dp[1][1] = (p1[0] == s[0]) ? 0 : -1000000000000000000LL;
      }
      for (long long j = 2; j <= (long long)p1.length(); j++) {
        dp[1][j] = -1000000000000000000LL;
      }
      for (long long i = 2; i <= (long long)s.length(); i++) {
        if (dp[i - 1][0] == (long long)p2.length()) {
          dp[i][0] = (long long)p2.length();
        } else {
          dp[i][0] = dp[i - 1][0] + ((p2[dp[i - 1][0]] == s[i - 1]) ? 1 : 0);
        }
        for (long long j = 1; j <= (long long)p1.length(); j++) {
          dp[i][j] = -1000000000000000000LL;
          if (s[i - 1] == p1[j - 1]) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
          }
          dp[i][j] = max(dp[i][j], dp[i - 1][j]);
          if (0 <= dp[i - 1][j] && dp[i - 1][j] < (long long)p2.length()) {
            if (p2[dp[i - 1][j]] == s[i - 1]) {
              dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
            }
          }
        }
      }
      if (dp[s.length()][p1.length()] >= (long long)p2.length()) {
        assert(dp[s.length()][p1.length()] == (long long)p2.length());
        possible = true;
        break;
      }
    }
    if (possible) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
