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)

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long 
#define inf 0x3f3f3f3f
#define infll 0x3f3f3f3f3f3f3f3f
#define all(x) (x).begin(),(x).end()
#define pb push_back

#define F_OR(i, a, b, s) for (int i=(a); (s)>0?i<(b):i>(b); i+=(s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...) F_ORC(__VA_ARGS__)(__VA_ARGS__)
#define EACH(x, a) for (auto& x: a)
 
// debug templates 
#define debug(x)       cerr<< #x <<" : "<< x << endl;
#define debuga(A,N)    cerr<< #A <<" : [";for(int i = 0; i<N;i++) cerr<<A[i]<<" "; cerr<<"]\n";
#define debuga2(A,N,M) cerr<< #A << " : \n"; for(int i=0;i<N;i++){cerr<<"[";for(int j=0;j<M;++j) cerr<<A[i][j]<<" ";cerr<<"]\n";}
#define debugp(p)      cerr<< #p <<" : "<<"("<<(p).first<<","<<(p).second<<")\n";
#define debugv(v)      cerr<< #v <<" : "<<"[";for(int i = 0; i< (v).size(); i++) cerr<<v[i]<<" "; cerr<<"]\n";
#define debugv2(v)     cerr<< #v << " : \n"; for(int i=0;i<v.size();i++){cerr<<"[";for(int j=0;j<(v[0].size());++j) cerr<<v[i][j]<<" ";cerr<<"]\n";}
#define debugs(m)      cerr<< #m <<" : [ "; for(auto itr = m.begin(); itr!=m.end();itr++) cerr<<*itr<<" "; cerr<<"]\n";
#define debugm(m)      cerr<< #m <<" : [ "; for(auto itr = m.begin();itr!=m.end(); itr++) cerr<<"("<<itr->first<<","<<itr->second<<") ";cerr<<"]\n";

int n,k; 

int solver(string s){
	
	// 0011...100 want index of last one in segment (1-indexed)
	vector<int> ones;
	char prev = '0';
	ones.pb(0);
	FOR(s.size()){
		if(s[i] == '0' && prev == '1'){
			ones.pb(i);
		}
		prev = s[i];
	}
	ones.pb(s.size());
	
	
	//debugv(ones);
	
	// if string starts with one first section of zeros hasnt occurred yet
	if(s[0] == '1'){
		ones.erase(ones.begin() + 1);
	}
	
	if(ones.size() - 1 <= k) return s.size();
	
	int ans = 0;
	int temp_k = k;
	FOR(i,1,ones.size()){
		if(temp_k) ans = ones[i], --temp_k;
		else{
			int temp_ans = ones[i] - ones[i-k];
			ans = max(temp_ans,ans);
		}
	}
	return ans;
}





void solve(){
	string s; cin >> n >> k >> s;
	string t = s;
	reverse(all(s));
	cout << max(solver(t),solver(s));
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	solve();
	return 0;
}