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)

/* Simplicity and Goodness */

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>

// using namespace __gnu_pbds;
using namespace std;
// typedef tree<int, null_type, less<int>, rb_tree_tag,
//             tree_order_statistics_node_update> indexed_set;

void my_dbg() { cout << endl; }
template<typename Arg, typename... Args> void my_dbg(Arg A, Args... B)
{ cout << ' ' << A; my_dbg(B...); }
#define dbg(...)  cout << "(" << #__VA_ARGS__ << "):", my_dbg(__VA_ARGS__)

#define scn(n) scanf("%d", &n)
#define lscn(n) scanf("%lld", &n)
#define pri(n) printf("%d ", (int)(n))
#define prin(n) printf("%d\n", (int)(n))
#define lpri(n) printf("%lld ", n)
#define lprin(n) printf("%lld\n", n)
#define rep(i,a,b) for(int i=(int)(a); i<(int)(b); i++)
#define pb push_back
#define mp make_pair
#define F first
#define S second

using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;

const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9+7;
const int N = 2e5+2;


void solve()
{
    int n, k;
    scn(n); scn(k);

    string s;
    cin >> s;

    s += '#';

    int cnt = 0;
    char prv = '1';

    vi util;

    rep(i, 0, s.size()) {
        if(s[i] != prv) {
            util.pb(cnt);
            cnt = 1;
        } else {
            cnt ++;
        }
        prv = s[i];
    }

    int ans = 0, sum = 0;

    rep(i, 0, util.size()) {
        sum += util[i];

        if(i - (2 * k + 1) >= 0) 
            sum -= util[i - (2 * k + 1)];

        if(i&1^1)
            ans = max(ans, sum);
    }

    prin(sum);
}

int main()
{
    int t = 1;
    // scn(t);
    
    while(t --) {
        solve();
    }
    return 0;
}