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
for _ in range(int(input())):
    s = input()
    t = input()
    if len(t) == 1:
        print("YES" if t in s else "NO")
        continue
    nxt = [[-1] * 26 for _ in range(len(s) + 1)]
    nxt[-2][ord(s[-1]) - ord('a')] = len(s) - 1
    for i in range(len(s) - 2, -1, -1):
        for c in range(26):
            nxt[i][c] = nxt[i + 1][c]
        nxt[i][ord(s[i]) - ord('a')] = i
    ans = "NO"
    for p in range(1, len(t)):
        a = t[:p]
        b = t[p:]
        dp = [[-1] * (len(b) + 1) for _ in range(len(a) + 1)]
        dp[0][0] = 0
        for la in range(len(a) + 1):
            for lb in range(len(b) + 1):
                if dp[la][lb] != -1:
                    if la < len(a):
                        if dp[la + 1][lb] != -1:
                            if nxt[dp[la][lb]][ord(a[la]) - ord('a')] != -1:
                                if nxt[dp[la][lb]][ord(a[la]) - ord('a')] < dp[la + 1][lb] - 1:
                                    dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
                                    dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
                        else:
                            dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
                            dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
                    if lb < len(b):
                        if dp[la][lb + 1] != -1:
                            if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] != -1:
                                if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] < dp[la][lb + 1] - 1:
                                    dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
                                    dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
                        else:
                            dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
                            dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
                if dp[len(a)][len(b)] != -1:
                    ans = "YES"
                    break
    print(ans)
