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 main():
    s = input()
    n = len(s)
    k = int(input())
    # dp[左端][左端からの右端の距離][使った回数]
    dp = [[[1]*(k+1) for _ in [0]*(n-i)] for i in range(n)]

    for i in range(n-1):
        if s[i] == s[i+1]:
            dp[i][1][0] = 2
    if k > 0:
        for i in range(n-1):
            dp[i][1][1] = 2

    for cnt in range(k):
        for d in range(2, n):
            for left in range(n-d):
                right = left+d
                plus = (s[left] == s[right])*2
                dp[left][d][cnt] = max(
                    dp[left][d][cnt], dp[left][d-1][cnt],
                    dp[left+1][d-1][cnt], dp[left+1][d-2][cnt]+plus)
                dp[left][d][cnt+1] = max(
                    max(dp[left][d][cnt:cnt+2]), dp[left+1][d-2][cnt]+2)
    for d in range(2, n):
        for left in range(n-d):
            right = left+d
            plus = (s[left] == s[right])*2
            dp[left][d][k] = max(
                dp[left][d][k], dp[left][d-1][k], dp[left+1][d-1][k], dp[left+1][d-2][k]+plus)
    print(dp[0][-1][-1])


main()