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

#define rep(i,n) for(ll i=0; i<(n); i++)
typedef long long ll;
using namespace std;

int main(){
    int n,k; cin>>n>>k;
    string s;
    cin>>s;
    
    vector<int> Nums;
    
    int now = 1;  //今見ている数
    int cnt = 0;//nowがいくつ並んでいるか

    rep(i, n){
        if(s[i] == (char)('0' + now)) cnt++;
        else{
            Nums.push_back(cnt);
            now ^=1;  //０と１を切り替えるときの計算
            cnt = 1;
        }
    }
    if(cnt!=0) Nums.push_back(cnt);
    
    
    if(Nums.size() % 2 == 0) Nums.push_back(0);
    
    int Add = 2*k+1;
    
    int ans = 0;
    
    for(int i=0; i < Nums.size(); i += 2){
        int tmp = 0;
        
        int left = i;
        int right = min((int)Nums.size(),i + Add);
        
        for(int j = left; j<right;j++){
            tmp += Nums[j];
        }
        ans = max(tmp,ans);
        
        
        
        
        
    }
    
    
    cout << ans <<endl;
    

    
    
}