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>
bool QQQQ;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, -1, 1, 1, -1, 1, -1};
using namespace std;
long long ttt = 1;
long long dp[410][410];
string s, t;
void solve() {
  cin >> s >> t;
  s = "!" + s;
  t = "!" + t;
  for (int i = 1; i < t.size(); i++) dp[0][i] = -1;
  for (int k = 1; k < t.size(); k++) {
    for (int i = 1; i < s.size(); i++)
      for (int j = 0; j <= min(i, k); j++) {
        dp[i][j] = -1;
        if (i > j && dp[i - 1][j] >= 0) {
          dp[i][j] = dp[i - 1][j];
          if (dp[i - 1][j] < t.size() - 1 - k &&
              t[k + dp[i - 1][j] + 1] == s[i])
            dp[i][j]++;
        }
        if (j && dp[i - 1][j - 1] >= 0 && s[i] == t[j])
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    if (dp[s.size() - 1][k] == t.size() - 1 - k) {
      puts("YES");
      return;
    }
  }
  puts("NO");
}
int main() {
  QQQQ = 0;
  scanf("%I64d", &ttt);
  for (int i = 1; i <= ttt; i++) solve();
  if (QQQQ) printf("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
  ;
}
