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
from sys import stdin, stdout

nxt = []

def find_it(s, left, right):
    global nxt
    dp = [[1000 for i in range(len(right) + 1)] for j in range(len(left) + 1)]
    dp[0][0] = 0
    for i in range(len(left)+1):
        for j in range(len(right)+1):
            if dp[i][j] > len(s):
                continue
            if j < len(right) and nxt[ord(right[j]) - 97][dp[i][j]] != -1:
                if nxt[ord(right[j]) - 97][dp[i][j]] < dp[i][j+1]:
                    dp[i][j+1] = nxt[ord(right[j]) - 97][dp[i][j]] +1
            if i < len(left) and nxt[ord(left[i]) - 97][dp[i][j]] != - 1:
                if nxt[ord(left[i]) - 97][dp[i][j]] < dp[i+1][j]:
                    dp[i+1][j] = nxt[ord(left[i]) - 97][dp[i][j]] +1
    if dp[len(left)][len(right)] != 1000:
        return True
    else:
        return False

def main():
    global nxt
    n = int(stdin.readline())
    for _ in range(n):
        s = stdin.readline().rstrip()
        t = stdin.readline().rstrip()
        nxt = [[-1 for _ in range(len(s)+1)] for i in range(26)] 
        for i,x in enumerate(s):
            nxt[ord(x) - 97][i] = i
        for i in range(26):
            for j in range(len(s)-1 ,-1,-1):
                if nxt[i][j] != j:
                    nxt[i][j] = nxt[i][j+1]
        
        r = False            
        for i in range(len(t)):
            res = find_it(s, t[:i], t[-len(t)+i:])
            if res is True:
                r = True
                break
        
        if r is True:
            stdout.write("YES\n")
        else:
            stdout.write("NO\n")
            

main()            