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 INF = 0x3f3f3f3f;
const int maxn = 400 + 5;
const int mod = 1e9 + 7;
string s, t;
int dp[maxn][maxn];
int check(string &t1, string &t2) {
  memset((dp), (-1), sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < s.length(); i++) {
    for (int j = 0; j <= t1.size(); j++) {
      if (dp[i][j] >= 0) {
        int cur = dp[i][j];
        if (j < t1.size() && s[i] == t1[j])
          dp[i + 1][j + 1] = max(dp[i + 1][j + 1], cur);
        if (cur < t2.size() && s[i] == t2[cur])
          dp[i + 1][j] = max(dp[i + 1][j], cur + 1);
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      }
    }
  }
  return dp[s.size()][t1.size()] == t2.size();
}
void solve() {
  cin >> s >> t;
  int len = t.size();
  for (int i = 0; i < len; i++) {
    string t1 = t.substr(0, i);
    string t2 = t.substr(i);
    if (check(t1, t2)) {
      printf("YES\n");
      return;
    }
  }
  printf("NO\n");
}
int main(int argc, char *argv[]) {
  if (argc == 2 && strcmp("-debug", argv[1]) == 0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}
