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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <climits>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <fstream>

using namespace std;
#define TOSTRING(x) #x
#define SZ(x) (int)(x).size()
#define REP(i,n) for(int i=0;i<(n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REPR(i,n) for(int i=(n)-1;i>=0;i--)
#define ALL(s) (s).begin(), (s).end()
#define so(V) sort(ALL(V))
#define rev(V) reverse(ALL(V))
#define uni(v) v.erase( unique(ALL(v)) , v.end());
#define PAU system("pause")

typedef long long unsigned int llu;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
const double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int INF = (1 << 28);
const ll LINF = 1e18;
const double PI = acos(-1);

template<typename T>vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T, typename... Ts>auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v<T>(ts...))>(a, make_v<T>(ts...)); }
template<typename T, typename U, typename... V>
typename enable_if<is_same<T, U>::value != 0>::type
fill_v(U& u, const V... v) { u = U(v...); }
template<typename T, typename U, typename... V>
typename enable_if<is_same<T, U>::value == 0>::type
fill_v(U& u, const V... v) { for (auto& e : u) fill_v<T>(e, v...); }
template<class T> bool chmax(T & a, const T & b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T & a, const T & b) { if (a > b) { a = b; return 1; } return 0; }

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, K;
	cin >> N >> K;
	string S;
	cin >> S;
	char c = '1';
	int cur = 0;
	vi V;
	queue<int> Q;
	REP(i, N) {
		if (S[i] == c) {
			cur++;
		}
		else {
			V.push_back(cur);
			cur = 1;
			c = S[i];
		}
	}
	V.push_back(cur);
	int ind = 0;
	int sum = 0;
	int ans = 0;
	while (ind < SZ(V)) {
		chmax(ans, sum + V[ind]);
		if (ind == SZ(V) - 1)break;
		sum += V[ind] + V[ind + 1];
		Q.push(V[ind] + V[ind + 1]);
		if (SZ(Q) > K) {
			sum -= Q.front();
			Q.pop();
		}
		ind += 2;
	}
	chmax(ans, sum);
	cout << ans << endl;
	return 0;
}