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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<ldouble, ldouble> pdd;

char S[310];
int K;
int D[310][310][310];

void solve() {
	scanf("%s%d", S+1, &K);
	int L = (int)strlen(S+1);
	for(int i=1;i<=L;i++) rep(j, K+1) D[i][i][j] = 1;
	for(int i=1;i<L;i++) for(int j=1;j+i<=L;j++) {
		int l = j, r = j + i;
		for(int k=0;k<=K+1;k++) {
			D[l][r][k] = max(D[l+1][r][k], D[l][r-1][k]);
			if(S[l] == S[r]) D[l][r][k] = max(D[l][r][k], D[l+1][r-1][k] + 2);
			else {
				if(k) D[l][r][k] = max(D[l][r][k], D[l+1][r-1][k-1] + 2);
			}
		}
	}
	printf("%d\n", D[1][L][K]);
}

int main(){
	int Tc = 1; //scanf("%d\n", &Tc);
	for(int tc=1;tc<=Tc;tc++){
		solve();
	}
	return 0;
}