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>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#ifdef leowang
#define debug(...) do{\
	fprintf(stderr,"%s - %d : (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
	_DO(__VA_ARGS__);\
}while(0)
template<typename I> void _DO(I&&x){cerr<<x<<endl;}
template<typename I,typename...T> void _DO(I&&x,T&&...tail){cerr<<x<<", ";_DO(tail...);}
#else
#define debug(...)
#endif
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
	out<<'('<<P.F<<','<<P.S<<')';
}
//}}}
const ll maxn=305;
const ll maxlg=__lg(maxn)+2;
const ll INF64=8000000000000000000LL;
const int INF=2000000000;
const ll MOD=ll(1e9+7);
const double PI=acos(-1);
//const ll p=880301;
//const ll P=31;
/*
ll mypow(ll a,ll b){
	ll res=1LL;
	while(b){
		if(b&1) a=a*a%MOD;
		else res=res*a%MOD;
		b>>=1;
	}
	res=res*a%mod;
	return res;
}
*/
string s,t;
int dp[maxn][maxn][maxn*2];

int f(int i,int j,int k){
	if(k<0) return -INF;
	if(i<=0||j<=0) return 0;
	if(dp[i][j][k]!=-1) return dp[i][j][k];
	if(s[i-1]==t[j-1]) return f(i-1,j-1,k)+1;
	return dp[i][j][k]=max({f(i-1,j,k),f(i,j-1,k),f(i-1,j-1,k-1)+(SZ(s)%2==0||i!=SZ(s)/2||j!=SZ(s)/2)});
}

int main()
{
	IOS;
	memset(dp,-1,sizeof dp);
	int k;
	cin>>s>>k;
	t=s;
	reverse(ALL(t));
	int ans=0;
	REP(i,SZ(s)){
		ans=max(ans,f(i,SZ(s)-1-i,k)*2+1);
	}
	REP(i,SZ(s)-1){
		if(s[i]==s[i+1]) ans=max(ans,2*f(i+1,SZ(s)-i-1,k));
		else ans=max(ans,2*f(i,SZ(s)-i-2,k-1)+2);
	}
	cout<<ans;
	return 0;
}