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
def main():
    T = int(input().strip())
    for _ in range(T):
        s = input().strip()
        t = input().strip()
        n = len(s)
        
        find = [[n] * 26 for _ in range(n + 2)]
        for i in range(n - 1, -1, -1):
            find[i][:] = find[i + 1]
            find[i][ord(s[i]) - ord("a")] = i

        def interleaving(a, b):
            dp = [n] * (len(b) + 1)
            for i in range(len(a) + 1):
                for j in range(len(b) + 1):
                    if i == j == 0:
                        dp[j] = -1
                        continue
                    res = n
                    if i > 0:
                        res = min(res, find[dp[j] + 1][ord(a[i - 1]) - ord("a")])
                    if j > 0:
                        res = min(res, find[dp[j - 1] + 1][ord(b[j - 1]) - ord("a")])
                    dp[j] = res
            return dp[-1] < n

        if any(interleaving(t[:i], t[i:]) for i in range(len(t))):
            print("YES")
        else:
            print("NO")


main()
