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<bits/stdc++.h>
using namespace std;
typedef long long ll;
# define mod 1000000007
# define INF (1 << 29)
# define MAX_N 1000010
/* ACはうれCすぎてうれD */


int main(void){ 
    int n,k;
    cin >> n >> k;
    string s;
    cin >> s;

    vector<int> v; // 0 か 1 の先頭のidxが入る
    v.push_back(0);
    for(int i=1; i<n; i++){
        if(s[i] != s[i-1]){
            v.push_back(i);
        }
    }
    v.push_back(n);
    /* for(int i=0; i<v.size(); i++){
        cout << v[i] << "\n";
    } */

    // 各 v の要素が先頭だとした時に
    // いくつの 1 を連続させることができるか数える
    // k 個 先をみることができる
    // 区間の長さは idx どうしの差として取得可能

    int ans = 0;
    for(int i=0; i<v.size(); i++){
        if(s[v[i]] == '0'){
            int tmp_ans = min(v[i + 2*k],n) - v[i];
            if(tmp_ans > ans) ans = tmp_ans;
        }
        else{
            int tmp_ans = min(v[i + 2*k + 1],n) - v[i];
            if(tmp_ans > ans) ans = tmp_ans;
        }
    }

    cout << ans;
    return 0;
}