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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>

using namespace std;

int csum[100005];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, k;
    string s;
    cin >> n >> k >> s;
    
    vector<int> sum;
    int now = 1;
    int id = 0;
    int cnt = 0;
    while(id < n){
        if(s[id] - '0' == now){
            ++cnt;
        }
        else{
            sum.push_back(cnt);
            now ^= 1;
            cnt = 1;
        }
        ++id;
    }
    sum.push_back(cnt);
    
    if(sum.size() % 2 == 0)     sum.push_back(0);
    
    // 累積和
    for(int i = 0; i < sum.size(); ++i){
        csum[i+1] = csum[i] + sum[i];
    }
    
    int ans = 0;
    int add = 2 * k + 1;
    for(int i = 0; i < sum.size(); i += 2){
        int left = i;
        int right = min(i + add, (int)sum.size());
        
        int temp = csum[right] - csum[left];
        
        ans = max(ans, temp);
    }
    
    
    cout << ans << endl;
}