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

T = int(input())

for loop in range(T):

    s = input()
    p = input()
    flag = False

    for i in range(len(p)):

        l = p[:i]
        r = p[i:]

        dp = [None] * (len(l) + 1)
        dp[0]= 0

        l += " "
        r += " "

        for S in s:

            for x in range(len(l)-1 , -1 , -1):

                #print (dp[x])

                if dp[x] != None and l[x] == S:

                    if dp[x+1] == None:
                        dp[x+1] = dp[x]
                    else:
                        dp[x+1] = max(dp[x+1] , dp[x])

                if dp[x] != None and r[dp[x]] == S:

                    dp[x] += 1

                #print (S,dp)

        #print (l,r,dp)

        if dp[-1] == len(r)-1:
            flag = True
            break

        

    if flag:
        print ("YES")
    else:
        print ("NO")