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
class NextStringIndex:
    def __init__(self, string):
        self.INF = 10 ** 9
        self.alph = "abcdefghijklmnopqrstuvwxyz"
        self.kind = len(self.alph)
        self.to_ind = {char: ind for ind, char in enumerate(self.alph)}

        self.string = string
        self.len_s = len(string)
        self.next_ = self.make_next()

    def __getitem__(self, tup):
        ind, char = tup
        return self.next_[ind][self.to_ind[char]]

    def make_next(self):
        dp = [[self.INF] * self.kind for i in range(self.len_s + 1)]
        for i in range(len_s)[::-1]:
            for j, char in enumerate(self.alph):
                if s[i] == char:
                    dp[i][j] = i + 1
                else:
                    dp[i][j] = dp[i + 1][j]
        return dp


def solve(t1, t2, len_s):
    INF = 10 ** 9
    len_t1 = len(t1)
    len_t2 = len(t2)
    dp = [[INF] * (len_t2 + 1) for i in range(len_t1 + 1)]
    dp[0][0] = 0

    for i in range(len_t1 + 1):
        for j in range(len_t2 + 1):
            length = dp[i][j]
            if length > len_s:
                continue
            if i < len_t1 and s_next[length, t1[i]] < INF:
                dp[i + 1][j] = min(dp[i + 1][j], s_next[length, t1[i]])
            if j < len_t2 and s_next[length, t2[j]] < INF:
                dp[i][j + 1] = min(dp[i][j + 1], s_next[length, t2[j]])

    return dp[-1][-1] < INF


query = int(input())
for _ in range(query):
    s = input()
    t = input()
    len_s = len(s)
    len_t = len(t)

    s_next = NextStringIndex(s)
    
    flag = False
    for i in range(len_t + 1):
        flag |= solve(t[0:i], t[i:], len_s)
    if flag:
        print("YES")
    else:
        print("NO")