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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <string.h>
#include <math.h>

int main (int argc, char const *argv[])
{
    int64_t N, K;
    char S[(int)1e5 + 1];
    int64_t len[(int)2e5] = {0}, tmp = 0, ans = 0;
    int64_t *p;

    scanf("%"PRId64"%"PRId64, &N, &K);
    scanf("%s", S);

    p = len;
    for (int i = 0; i < N; i++) {
        if (S[i] == S[i + 1]) {
            *p += 1;
        } else {
            *p += 1;
            p++;
        }
    }

    p = len;
    for (int i = 0; i < (K * 2); i++) {
        tmp += *p++;
        if (*p == 0) break;
    }
    if (S[0] == '1') tmp += *p;

    ans = tmp;

    fprintf(stderr, "%"PRId64"\n", tmp);

    if (S[0] == '1') {
        for (int i = 1; len[i] != 0; i++) {
            tmp -= len[i - 1];
            fprintf(stderr, "-%"PRId64, len[i - 1]);
            if ((i % 2) == 0) {
                tmp += len[i + K * 2 - 1];
                tmp += len[i + K * 2];
                fprintf(stderr, "+%"PRId64, len[i + K * 2 - 1]);
                fprintf(stderr, "+%"PRId64, len[i + K * 2]);
            }
            fprintf(stderr, " %"PRId64"\n", tmp);
            if (tmp > ans) ans = tmp;
        }
    } else {
        fprintf(stderr, "zero\n");
        for (int i = 1; len[i] != 0; i++) {
            tmp -= len[i - 1];
            fprintf(stderr, "-%"PRId64, len[i - 1]);
            if ((i % 2) == 1) {
                tmp += len[i + K * 2 - 1];
                tmp += len[i + K * 2];
                fprintf(stderr, "+%"PRId64, len[i + K * 2 - 1]);
                fprintf(stderr, "+%"PRId64, len[i + K * 2]);
            }
            fprintf(stderr, " %"PRId64"\n", tmp);
            if (tmp > ans) ans = tmp;
        }
    }

    printf("%"PRId64"\n", ans);
}