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 INF 1001000100010001000
#define MOD 1000000007
#define EPS 1e-10
#define int long long
#define rep(i, N) for (int i = 0; i < N; i++)
#define Rep(i, N) for (int i = 1; i < N; i++)
#define For(i, a, b) for (int i = (a); i < (b); i++)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define pii pair<int, int>
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define vp vector< pii >
#define all(a) (a).begin(), (a).end()
#define Int(x) int x; cin >> x;
#define int2(x, y) Int(x); Int(y);
#define int3(x, y, z) Int(x); int2(y, z);
#define in(x, a, b) ((a) <= (x) && (x) < (b))
#define fir first
#define sec second
#define ffir first.first
#define fsec first.second
#define sfir second.first
#define ssec second.second
#define Decimal fixed << setprecision(10)

//int dxy[5] = {0, 1, 0, -1, 0};
// cmd

signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    int2(n, k);
    string st;
    cin >> st;
    
    vp mmnmm;
    int mm = 0;
    char state = 0;
    rep(i, st.size()) {
        if (state == 0) {
            state = st[i];
            mm++;
        } else if (state != st[i]) {
            mmnmm.pb({mm, (int)(state == '1')});
            mm = 1;
            state = st[i];
        } else {
            mm++;
        }
    }
    mmnmm.pb({mm, (int)(state == '1')});

    int ans = 0, cnt = 0, ret = 0, yui = 0;
    rep(i, mmnmm.size()) {
        if (cnt < k) {
            ans += mmnmm[i].fir;
            if (!mmnmm[i].sec) cnt++;
        } else if (cnt == k) {
            ans += mmnmm[i].fir;
            if (!mmnmm[i].sec) {
                cnt++;
                while (cnt != k) {
                    ans -= mmnmm[yui].fir;
                    cnt -= !mmnmm[yui].sec;
                    yui++;
                }
            }
        }
        //cout << cnt << " " << ans << endl;
        ret = max(ret, ans);
    }
    std::cout << ret << std::endl;

    return 0;
}