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: Runtime Error)

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

using namespace std;

int main(){
    int N;
    int K;
    cin >> N >> K;
    string S;
    cin >> S;
    vector<int> nums;
    int counter = 0;
    int now = 1;
    for(int i = 0; i < N; i++){
        if(S[i] == (char)(now + '0')) counter++;
        else{
            nums.push_back(counter);
            now = 1 - now;
            counter = 1;
        }
    }
    if(counter != 0) nums.push_back(counter);
    if(now == '0') nums.push_back(0);//最後が0カウンタならば架空の1の個数0を付け足す。1-0-1-0-1とかにしたい

    vector<int> sums(nums.size()+1,0);
    int sum = 0;
    for (int i = 0; i < sums.size(); i++)
    {
        sums[i + 1] = sums[i]+nums[i];
    }


    int ans;
    int left = 0;
    int right = 0;
    int Add = 2*K+1;
    int temp;
    for (int i = 0; i < sums.size(); i+=2)
    {
        left = i;
        right = min(left + Add, (int)nums.size());
        temp = sums[right] - sums[left];
        ans = max(ans,temp);
    }
    cout << ans << endl;
}