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/env python3
import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    s = list(input().rstrip())
    t = input().rstrip()
    ok = False
    for i in range(len(t)):
        t1 = list(t[:i]) + ["#"]
        t2 = list(t[i:]) + ["#"] 
        # dp[seen i-th index][match j in front] = match in back
        dp = [[-1] * (len(t) + 1) for _ in range(len(s) + 1)]
        dp[0][0] = 0
        for j, ch in enumerate(s):
            for k in range(len(t1)):
                if dp[j][k] == -1:
                    continue
                dp[j+1][k] = max(dp[j+1][k], dp[j][k])
                if ch == t1[k]:
                    dp[j+1][k+1] = max(dp[j+1][k+1], dp[j][k])
                if ch == t2[dp[j][k]]:
                    dp[j+1][k] = max(dp[j+1][k], dp[j][k] + 1)
        for k in range(len(t) + 1):
            if dp[len(s)][k] + k >= len(t):
                ok = True

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