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 int Maxn = 405;
const int Inf = 1000000000;
int T;
string s, t;
int dp[Maxn];
bool Solve(int x) {
  fill(dp, dp + Maxn, -Inf);
  dp[x] = 0;
  for (int i = 0; i < s.length(); i++)
    for (int j = int(t.length()); j >= x; j--)
      if (dp[j] >= 0) {
        if (j + 1 <= t.length() && t[j] == s[i])
          dp[j + 1] = max(dp[j + 1], dp[j]);
        if (dp[j] < x && s[i] == t[dp[j]]) dp[j]++;
      }
  return dp[t.length()] >= x;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    cin >> s;
    cin >> t;
    bool ok = false;
    for (int i = 0; i < t.length() && !ok; i++) ok = Solve(i);
    printf("%s\n", ok ? "YES" : "NO");
  }
  return 0;
}
