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;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define REP(i, n) for(int i=0; i<(n); ++i)
#define FOR(i, a, b) for(int i=(a); i<(b); ++i)
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)

#define DEBUG(x) cout<<#x<<": "<<(x)<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";REP(i, v.size())cout<<' '<<v[i];cout<<'\n'
#define ALL(a) (a).begin(), (a).end()

template<typename T> inline void CHMAX(T& a, const T b) {if(a<b) a=b;}
template<typename T> inline void CHMIN(T& a, const T b) {if(a>b) a=b;}

const ll MOD=1000000007ll;
// const ll MOD=998244353ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

const double EPS=1e-11;
#define EQ0(x) (abs((x))<EPS)
#define EQ(a, b) (abs((a)-(b))<EPS)

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	// cout<<setprecision(10)<<fixed;
	int n, k;
	string s;
	cin>>n>>k>>s;
	int r=0;
	while(r<n && s[r]=='1'){
		++r;
	}
	REP(i, k){
		while(r<n && s[r]=='0'){
			++r;
		}
		while(r<n && s[r]=='1'){
			++r;
		}
	}
	if(r==n){
		cout<<n<<'\n';
		return 0;
	}
	int ans=r, l=0;
	while(l<n){
		while(l<n && s[l]=='1'){
			++l;
		}
		while(l<n && s[l]=='0'){
			++l;
		}
		while(r<n && s[r]=='0'){
			++r;
		}
		while(r<n && s[r]=='1'){
			++r;
		}
		CHMAX(ans, r-l);
	}
	cout<<ans<<'\n';
	return 0;
}