Problem:
Score : 900 points

Problem Statement

Takahashi has decided to give a string to his mother.

The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T.
That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).

Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value.
Find the highest possible value achievable.

Constraints

1 \leq |S| \leq 300

0 \leq K \leq |S|

S consists of lowercase English letters.

K is an integer.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the highest possible value achievable.

Sample Input 1

abcabcabc
1

Sample Output 1

7

Changing the first character to c results in cbcabcabc.
Let this tring be T, then one longest common subsequence of T and T' is cbabcbc, whose length is 7.

Sample Input 2

atcodergrandcontest
3

Sample Output 2

15

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

def solve(s, k):
    n = len(s)
    if k >= n // 2:
        return n
    pdp = [[0] * (k + 1)]
    for l in range(1, n + 1):
        i = n - l
        dp = [[0] * (k + 1) for _ in range(l + 1)]
        dpr1, pdpr1 = dp[-1], [0] * (k + 1)
        for r in range(l):
            dpr, pdpr = dp[r], pdp[r - 1]
            inc = 2 if r else 1
            if s[i] == s[i + r]:
                for p in range(k + 1):
                    dp[r] = [c + inc for c in pdpr1]
            else:
                dpr[0] = max(pdpr[0], dpr1[0])
                for p in range(1, k + 1):
                    dpr[p] = max(pdpr[p], dpr1[p], pdpr1[p - 1] + inc)
            dpr1, pdpr1 = dpr, pdpr
        pdp = dp

    return pdp[-2][k]


print(solve(input(), int(input())))