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
#!/usr/bin/python3
# @Author  : indiewar

def check(s,t1,t2):
    s1 = len(t1)
    s2 = len(t2)
    n = len(s)
    dp = [[-1] * (s1+1) for i in range(n + 1)]
    dp[0][0] = 0
    for i in range(n):
        for j in range(s1+1):
            if dp[i][j] >= 0:
                if  j < s1 and t1[j] == s[i]:
                    dp[i+1][j+1] = max(dp[i+1][j+1],dp[i][j])
                if dp[i][j] < s2 and t2[dp[i][j]] == s[i]:
                    dp[i+1][j] = max(dp[i+1][j],dp[i][j]+1)
            dp[i+1][j] = max(dp[i+1][j],dp[i][j])

    # print(dp[n][s1])
    if dp[n][s1] == s2:
        return True
    else:
        return False


def solve():
    s = input()
    t = input()
    le = len(t)

    for i in range(le):
        t1 = t[:i]
        t2 = t[i:]
        if check(s,t1,t2) == True:
            print("YES")
            return
    print("NO")


T = int(input())
while T:
    T -= 1
    solve()