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 ti in range(T):
    s, t = input().strip(), input().strip()
    N = len(t)
    for i in range(1, N+1):
        # t1 : [0,i), t2 : [i,N)に分割
        dp = [0]+[-1]*i
        for l, c in enumerate(s):
            for j in range(i, -1, -1):
                # t1のprefix j を構成した時にt2はどこまで構成できるか
                tmp = dp[j]
                # cをt2に使う
                if dp[j] != -1 and i+dp[j] < N and t[i+dp[j]] == c:
                    tmp = dp[j]+1
                # cをt1に使う
                if j != 0 and t[j-1] == c:
                    tmp = max(tmp, dp[j-1])
                dp[j] = tmp
        if dp[i] == N-i:
            print("YES")
            break
    else:
        print("NO")

