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
import sys

readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')


def solve():
    s = list(map(lambda x: x-97, ns()))
    t = list(map(lambda x: x-97, ns()))
    n, m = len(s), len(t)
    nxt = [[n+1]*26 for _ in range(n+2)]
    for i in range(n-1, -1, -1):
        nxt[i] = nxt[i+1][:]
        nxt[i][s[i]] = i
    for b in range(m):
        t1 = t[:b]
        t2 = t[b:]
        dp = [[n+1]*(m-b+1) for _ in range(b+1)]
        dp[0][0] = 0
        for j in range(b+1):
            for k in range(m-b+1):
                if j:
                    dp[j][k] = min(dp[j][k], nxt[dp[j-1][k]][t1[j-1]] + 1)
                if k:
                    dp[j][k] = min(dp[j][k], nxt[dp[j][k-1]][t2[k-1]] + 1)
        # print(s, t1, t2)
        # prn(dp)
        if dp[b][m-b] <= n:
            print('YES')
            return
    print('NO')
    return

# solve()

T = ni()
for _ in range(T):
    solve()
