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 <string.h>
#include <math.h>
#include <stdlib.h>
#define MOD 1000000007
#define MAX(a,b) (a>b?a:b)
#define MAX3(a,b,c) MAX(a,MAX(b,c))
#define MIN(a,b) (a<b?a:b)
#define MIN3(a,b,c) MIN(a, MIN(b,c))
typedef long long lli;
int cmp(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}

int main()
{
    int i, n, k, state, len[100010]={0};
    char s[100010];
    scanf("%d %d",&n, &k);
    scanf("%s",s);
    // ex: 00011000011011 => len[]={0, 3, 2, 4, 2, 1, 2, 0, 0, ..., 0}
    int t = 1;
    for(i=0; i<n; i++) {
        if(s[i]=='0') {
            if(state==1) t++;
            len[t]++;
            state = 0;
        } else {
            if(state==0) t++;
            len[t]++;
            state = 1;
        }
    }
    int m = t;
    //for(i=0; i<=m+2; i++) printf("len[%d]=%d\n", i, len[i]);
    int temp=0;
    int v = (s[0]=='0'?0:1);
    for(i=v; i<=2*k+v && i<=m; i++) {
        temp += len[i];
    }
    int ans = temp;
    //printf("temp[0]=%d\n", temp);
    // 1d DP
    for(i=2*k+v; i<=m; i+=2) {
        temp = temp + len[i+1] + len[i+2] - len[i-2*k] - len[i-2*k+1];
        ans = MAX(ans, temp);
        //printf("%d %d %d %d  ",len[i+1], len[i+2], -len[i-2*k], -len[i-2*k+1]);
        //printf("temp[%d]=%d\n",i, temp);
    }
    printf("%d",ans);
    return 0;
}