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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

int main() {
    string s;
    int change;
    cin >> s >> change;
    int n = s.size();

    string t;
    for (int i = n - 1; i >= 0; i--)t += s[i];

    vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(n + 1, vector<int>(change + 1, -114514)));
    for (int i = 0; i <= n; i++) {
        for (int k = 0; k <= change; k++) {
            dp[i][0][k] = 0;
            dp[0][i][k] = 0;
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 0; k <= change; k++) {
                if (s[i - 1] == t[j - 1])dp[i][j][k] = dp[i - 1][j - 1][k] + 1;
                else dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k]);

                if (k > 0) {
                    dp[i][j][k] = max({dp[i][j][k - 1],
                                       dp[i - 1][j][k - 1] + 1,
                                       dp[i][j - 1][k - 1] + 1});
                }
            }
        }
    }

    cout << dp[n][n][change] + change << endl;
    return 0;
}