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: Accepted)

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

    return pdp[-2][k]


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