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 <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>

#define REP(i,n) for(i=0;i<n;i++)

int iin() {
    int i;
    scanf("%d", &i);
    return i;
}

long long llin() {
    long long i;
    scanf("%lld", &i);
    return i;
}

double din() {
    double i;
    scanf("%lf", &i);
    return i;
}

char cin() {
    char c;
    scanf("%c", &c);
    return c;
}

void stin(char *s) {
    scanf("%s", s);
}

void iout(int i) {
    printf("%d\n", i);
}

void llout(long long i) {
    printf("%lld\n", i);
}

void dout(double i) {
    printf("%lf\n", i);
}

void cout(char c) {
    printf("%c\n", c);
}

void stout(char *s) {
    printf("%s\n", s);
}

int main() {
    int n = iin(), k = iin();
    char s[n];
    stin(s);
    int t[n];
    memset(t, 0, sizeof t);
    int i, index = 0, first, c, count = 0, last;
    first = s[0] - '0';
    last = s[n - 1] - '0';
    c = first;
    REP(i, n) {
        if (s[i] != (c + '0')) {
            if (s[i] == '1') count++;
            if (index > 0) t[index] += t[index - 1];
            index++;
            c != c;
        }
        t[index]++;
    }
    if (index > 0) t[index] += t[index - 1];

    if (last == 0) count++;

    if (count <= k) {
        iout(n);
        return 0;
    }

    int max = 0, tmp;
    for (i = -1; i < index; i+=2) {
        if (first == 0) {
            if (i == -1) {
                tmp = t[k * 2 - 1];
            } else {
                if (k * 2 + 1 + i < index) {
                    tmp = t[k * 2 + 1 + i] - t[i];
                } else if (k * 2 + 1 + i == index) {
                    tmp = t[k * 2 + i] - t[i];
                } else {
                    break;
                }
            }
        } else {
            if (i == -1) {
                tmp = t[k * 2 + 1 + i];
            } else {
                if (k * 2 + 1 + i == index) {
                    tmp = t[k * 2 + i] - t[i];
                } else if (k * 2 + 1 + i < index) {
                    tmp = t[k * 2 + 1 + i] - t[i];
                } else {
                    break;
                }
            }
        }
        if (max < tmp) {
            max = tmp;
        }
    }
    iout(max);
    return 0;
}