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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#define REP(i,a) for (int i = 0;i < (a);++i)
#define FOR(i,a,b) for (int i = (a);i < (b); ++i)
#define FORR(i,a,b) for (int i = (a);i >= (b); --i)
#define ALL(obj) (obj).begin,(obj).end()
#define SORT(list) sort(ALL((list)));
using namespace std;
using ll = long long;
int main(){
    int n,k;
    cin >> n >> k;
    string s;
    cin >> s;
    int count=0;
    int r=0;
    int answer=0;
    if(s[0]=='0')count++;
    REP(l,n){
        while(count <= k){
            if(r>=n-1){
                break;
            }
            r++;
            if(s[r]!=s[r+1]){
                if(s[r]=='1'&&(r==n-1||s[r+1]=='0')){
                    count++;
                    while(r!=n-1&&s[r+1]=='1'){
                        //if(r>=n-1)break;
                        r++;
                    }
                }
            }
        }
        if(answer < r-l+1){
            answer = r-l+1;
        }
         //cout << l << " " << r << " " << count << endl;
        //if(r!=n-1&&s[r]=='1')continue;
        
        if(l!=0&&s[l] != s[l+1]&&s[l]=='0'){
            count--;
        }
       
    }
    cout << answer << endl;
    return 0;
}