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;
using ll = long long;
using ld = long double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define reps(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
const ll mod = ll(1e9) + 7;
const int INF = int(1e9);
int main()
{
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    bool flag = true;
    if (S[0] == '0')
    {
        flag = false;
    }
    int cnt = 0;
    vector<pair<int, int>> x;
    rep(i, S.size())
    {
        if (S[i] == '1' && !flag)
        {
            x.push_back({cnt, 1});
            flag = !flag;
            cnt = 0;
        }
        if (S[i] == '0' && flag)
        {
            x.push_back({cnt, 0});
            flag = !flag;
            cnt = 0;
        }
        cnt++;
    }
    x.push_back({cnt, (flag ? 0 : 1)});
    vector<int> k;
    vector<int> sum;
    k.push_back(0);
    sum.push_back(0);
    int sum_ = 0, k_ = 0;
    rep(i, x.size())
    {
        sum_ += x[i].first;
        k_ += x[i].second;
        sum.push_back(sum_);
        k.push_back(k_);
    }
    vector<int> sum_max(k_ + 1), sum_min(k_ + 1);
    rep(i, sum_min.size()) sum_min[i] = INF;
    rep(i, sum.size())
    {
        sum_min[k[i]] = min(sum_min[k[i]], sum[i]);
        sum_max[k[i]] = max(sum_max[k[i]], sum[i]);
    }
    int ans = 0;
    rep(i, k_ + 1)
    {
        ans = max(ans, sum_max[min(i + K, int(sum_max.size()) - 1)] - sum_min[i]);
    }
    rep(i, sum_max.size())
    {
        //cout << sum_max[i] << " ";
    }
    cout << ans << endl;

    return 0;
}