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 <bits/stdc++.h>
using namespace std;
#define IO ios_base::sync_with_stdio(false); cin.tie(NULL);

#define endl '\n'
#define D(x) cout << #x << " = " << (x) << endl;

const int maxn = 301;
int dp[maxn][maxn][maxn];

int f(string &s, int i, int j, int k) {
  if (i > j or i >= s.size() or j < 0) return 0;

  if (dp[i][j][k] == -1) {
    int ret = 0;
    if (s[i] == s[j]) {
      int cnt = 1 + (i != j) + f(s, i + 1, j - 1, k);
      ret = max(ret, cnt);
    } else {
      int cnt = f(s, i + 1, j, k);
      ret = max(ret, cnt);

      cnt = f(s, i, j - 1, k);
      ret = max(ret, cnt);

      if (k) {
        cnt = 2 + f(s, i + 1, j - 1, k - 1);
        ret = max(ret, cnt);
      }
    }

    dp[i][j][k] = ret;
  }

  return dp[i][j][k];
}

int main() { IO;
  string s;
  cin >> s;

  int k;
  cin >> k;

  memset(dp, -1, sizeof dp);
  int ans = f(s, 0, s.size() - 1, k);
  cout << ans << endl;

  return 0;
}