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 problem(s, p):
    n = len(s)
    F = [[n] * 26 for _ in range(n + 2)]
    for i in range(n - 1, -1, -1):
        F[i][:] = F[i + 1]
        F[i][ord(s[i]) - 97] = i

    def interleaving(l, r):
        dp = [-1] + [n] * len(r)

        for j in range(1, len(r) + 1):
            dp[j] = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]

        for i in range(1, len(l) + 1):
            dp[0] = F[dp[0] + 1][ord(l[i - 1]) - 97]

            for j in range(1, len(r) + 1):
                a = F[dp[j] + 1][ord(l[i - 1]) - 97]
                b = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]
                dp[j] = min(a, b)

        return dp[-1] < n

    for i in range(len(p)):
        if interleaving(p[:i], p[i:]):
            return 'YES'
    return 'NO'


for _ in range(int(input())):
    print(problem(input(), input()))
