Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdint.h>

#define STR_MAX    (100000)

#define FAST_TRIAL  (1)

uint64_t serialNum[STR_MAX+2];

int main(void)
{
    uint64_t n;
    uint64_t k;
    char s[STR_MAX+1];

    scanf("%ld %ld", &n, &k);
    scanf("%s", s);

    uint64_t ptr = 0;
    char curC;
    char nxtC;

    uint64_t indx   = 0;
    uint64_t head   = 0;
    uint64_t max    = 0;
    uint64_t cntK   = 0;
    uint64_t curSum = 0;

    curC = s[ptr++];

    while ( (curC != '0') ) {
        nxtC = s[ptr++];

        serialNum[indx]++;

        if ( (nxtC == '\0') ||
             ((curC != nxtC) && curC == '0') ) {
            /* 0が終わったので curSum に加算 */
            curSum += serialNum[indx++];

            /* 今, 逆立ち可能な最大数を増やす */
            cntK++;

            /* 逆立ち可能な最大数が k より大きい */
            if ( cntK > k ) {
                /* 先ほど加算したことで curSum が k+1 個になっている */
                curSum -= serialNum[head++];
            }
            /* ここまでで今回のserialNumによるcurSumが決定するので最大値更新 */
            max = ( max < curSum ) ? curSum : max;
        }

        curC = nxtC;
    }

    printf("%ld\n", max);

    return 0;
}