CQUOJ 10672 Kolya and Tandem Repeat

A. Kolya and Tandem Repeat

Time Limit: 2000ms

Memory Limit: 262144KB

64-bit integer IO format: %I64d      Java class name: Any

Submit Status PID: 10672

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 Input

Input

aaba
2

Output

6

Input

aaabbbb
2

Output

6

Input

abracadabra
10

Output

20

Hint

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

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

 

/*
2016年4月22日00:01:53

题意：给出一个字符串，给出k,可以向该字符串尾部添加k个字符串，求最长的连续重复两次的子串

    看了题解，可以先把这k个字符填成'*',再暴力枚举起点和长度，找出最大的长度
*/

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# define INF 0x3f3f3f3f
using namespace std;

int main(void)
{
    char s[20005];
    int n, len, i, j, k, flag, ans;
    while (~scanf("%s %d", &s, &n)){
        getchar();
        len = strlen(s);
        for (i = len; i < len+n; i++)
            s[i] = '*';
        len += n;
        ans = -1;
        for (i = 0; i < len; i++){
            for (j = 1; 2*j+i<=len; j++){
                flag = 1;
                for (k = i; k < i+j; k++){
                    if (s[k]!=s[k+j] && s[k+j]!='*'){
                        flag = 0;
                        break;
                    }
                }
                if (flag) ans = max(ans, 2*j);
            }
        }    
        printf("%d
", ans);
    }

    return 0;    
}