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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

// https://atcoder.jp/contests/abc124/tasks/abc124_d
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define CEIL(a, b) ((a - 1) / b + 1)

void solve() {
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    // 同じ数が連続したら1つとして扱う
    int seq = 0;
    char before = S[0];
    vector<pair<char, int>> v;
    S += ' ';
    REP(i, S.size()) {
        if (S[i] != before) {
            v.emplace_back(before, seq);
            seq = 0;
        }
        seq++;
        before = S[i];
    }
    // 0,1,0,1...を探す
    int cnt0 = 0, cnt1 = 0, sum = 0, ans = 0;
    queue<int> st;
    REP(i, v.size()) {
        sum += v[i].second;
        st.push(v[i].second);
        if (v[i].first == '0') {
            cnt0++;
            if (cnt0 > K) {
                cnt0--;
                sum -= st.front();
                st.pop();
            }
        } else {
            cnt1++;
            // 1,0,1,0,1 は許可するため1個多い
            if (cnt1 > K + 1) {
                cnt1--;
                sum -= st.front();
                st.pop();
            }
        }
        ans = max(ans, sum);
    }
    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}