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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>

using namespace std;


int main(){
    int N,K;
    cin >> N >> K;
    string S;
    cin >> S;

    vector<int> Nums;

    //1-0-1-0-1-0を作りたい。
    int now = 1;
    int cnt = 0;
    for(int i = 0; i < N; i++){
        if(S[i] == (char)('0'+now)) cnt++;
        else{
            Nums.push_back(cnt);
            //０と1を切り替える。
            now = 1 - now;
            cnt = 1;
        }
    }
    //最後が加えられてなかったら加える
    if(cnt != 0) Nums.push_back(cnt);

    //1-0-1-0のようにするには偶数にならないといけない。偶数じゃなかったら最後の0が足りてないってことだから0を足してあげる。
    if(Nums.size() % 2 == 0) Nums.push_back(0);

    //今回の見る範囲
    int Add = 2 * K + 1;

    //事前に変化していく変数を用意する。
    int left = 0;
    int right = 0;
    int tmp = 0; //leftからrightを含めないまでの合計

    int ans = 0;

    for(int i = 0; i < Nums.size(); i += 2){
        //移動していくゴール地点を用意。
        int Nextleft = i;
        int Nextright = min(i + Add, (int)Nums.size());

        //ゴール地点に行くまで、左は減らしてみぎは足していく。
        while(Nextleft > left){
            tmp -= Nums[left];
            left++;
        }
        while(Nextright > right){
            tmp += Nums[right];
            right++;
        }

        ans = max(ans, tmp);
    }

    cout << ans << endl;
    return 0;
}