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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int main(){
    string s;
    int n,k,cnt = 0,cnt1=0;
    vector<int> a;
    cin >> n >> k >> s;
    rep(i,n){
        int j = i;
        if(s[j]=='0') cnt1++;
        while(s[j+1]==s[j]&&j+1<n) j++;
        a.push_back(j-i+1);
        i = j;
    }
    int h = a.size();
    if(k>cnt1) k=cnt1;
    vector<int> b(h+1);
    rep(i,h) b[i+1] = b[i] + a[i];
    for(int i=0;i<h+1;i+=2){
        if(i%2==(s[0]=='1')) cnt = max(cnt,b[min(i+2*k,h)]-b[i]);
        else cnt = max(cnt,b[min(i+2*k+1,h)]-b[i]);
    }
    cout << cnt << endl;
}