cf 443B Kolya and Tandem Repeat

B. Kolya and Tandem Repeat

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kolya got string s for his birthday, the string consists of small English letters. He immediately added k more characters to the right of the string.

Then Borya came and said that the new string contained a tandem repeat of length l as a substring. How large could l be?

See notes for definition of a tandem repeat.

Input

The first line contains s (1 ≤ |s| ≤ 200). This string contains only small English letters. The second line contains number k (1 ≤ k ≤ 200) — the number of the added characters.

Output

Print a single number — the maximum length of the tandem repeat that could have occurred in the new string.

Sample test(s)

input

aaba
2

output

6

input

aaabbbb
2

output

6

input

abracadabra
10

output

20

Note

A tandem repeat of length 2n is string s, where for any position i (1 ≤ i ≤ n) the following condition fulfills: si = si + n.

In the first sample Kolya could obtain a string aabaab, in the second — aaabbbbbb, in the third — abracadabrabracadabra.

 

大力出奇迹2333

 

/*************************************************************************
    > File Name: code/2015summer/#3/A.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年07月28日 星期二 12时27分08秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int N=1E4+5;
char a[N];
int k;
int main()
{

    cin>>a>>k;
    int l=strlen(a);
    int m,ans;
    m=l+k;
    if (m%2==1) m--;
    if(k>=l)
    {
    cout<<m<<endl;
    return 0;
    }
    int max=0;
    for(int i = 0 ; i < l ; i++)
    {
        for(int j = 1 ; j <= l-i;j++)
        {
        ans = 0;
            for(int o = i ; o < i+j ; o++)
            {
        
                if(o+j>=l&&o+j<l+k)
                    ans++;
                else if(a[o]==a[o+j])
                     ans++;
            }
            if(ans==j&&2*ans>max)
              max=2*ans;
        }
    }
    cout<<max<<endl;
    return 0;
}