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<bits/stdc++.h>
#define pb           push_back
#define pii          pair<int,int>
#define vi           vector<int>
#define vii         vector<pair<int,int>>
#define mp           make_pair
#define FOR(i, n) for (int i = 0; i <  (int)(n); i++)
#define FOR1(i, n) for (int i = 1; i <= (int)(n); i++)
#define FORD0(i, n) for (int i = (int)n; i >= 0; i--)
#define FORD1(i, n) for (int i = (int)n; i >= 1; i--)
#define debug(X) { cout << #X << " = " << (X) << endl; }
#define endl '\n'
#define int long long int
using namespace std;


signed main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);

    string s;cin >> s;
    int n = s.length();
    int k;cin >> k;
    vector<vi> dp(n+1, vi(n+1));
    vector<vi> prev(n+1, vi(n+1));
    for(int change = 0; change <= k; change++){
        for(int len = 1; len <= n; len++){
            for(int start =0; start <= n - len; start++){
                dp[start][len] = max(dp[start+1][len-1], dp[start][len-1]); 
                if(s[start] == s[start + len -1]){
                    dp[start][len] = 1 + (len != 1) + dp[start + 1][len - 1];
                }else if(change > 0)
                    dp[start][len] = max(1 + (len != 1) + prev[start+1][len-1], dp[start][len]); 
            }
        }
        swap(dp, prev);
    }
    cout << prev[0][n];
}