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 <bits/stdc++.h>
using namespace std;

#define ALL(v) v.begin(), v.end()
#define V vector
#define P pair
typedef long long ll;
template<class T> void chmax(T& a, T b) { if(a < b) a = b; }

vector<pair<char, int>> runlength(string s) {
    int n = s.size();
    vector<pair<char, int> > res;
    char pre = s[0];
    int cnt = 1;
    for(int i = 1; i < n; i++){
        if(pre != s[i]){
            res.push_back({ pre, cnt });
            pre = s[i];
            cnt = 1;
        }
        else cnt++;
    }
    res.push_back({pre, cnt});
    return res;
}

int main() {
    int n, k; cin >> n >> k;
    string s; cin >> s;
    auto run = runlength(s);

    int m = run.size();
    int ans = 0, sum = 0, l = 0, z = 0;
    for(int r = 0; r < m; r++){
        sum += run[r].second;
        if(run[r].first == '0') z++;
        while(k < z){
            sum -= run[l].second;
            if(run[l].first == '0') z--;
            l++;
        }
        chmax(ans, sum);
    }
    cout << ans << endl;

}