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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef pair<P,int> T;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 2000000000
#define sz(x) ((int)(x).size())
#define fi first
#define sec second
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define repn(i,a,n) for(int (i)=(a);(i)<(int)(n);(i)++)
#define EQ(a,b) (abs((a)-(b))<eps)
template<class T> void chmin(T& a,const T& b){if(a>b)a=b;}
template<class T> void chmax(T& a,const T& b){if(a<b)a=b;}
int N,K;
string s;
deque<P> q;
void push_back(int v){
	if(q.empty())q.push_back(P(v,1));
	else if(q.back().fi==v)q.back().sec++;
	else q.push_back(P(v,1));
}
void pop_front(){
	assert(!q.empty());
	if(q.front().sec==1)q.pop_front();
	else q.front().sec--;
}
bool check(int k){
	for(int i=0;i<N;i++){
		if(i>=k)pop_front();
		push_back(s[i]-'0');
		if(i>=k-1){
			int sz = q.size();
			if(q.front().fi==1)sz--;
			if(sz!=0&&q.back().fi==1)sz--;
			assert(sz%2==1);
			if((sz+1)/2<=K)return true;
		}
	}
	return false;
}
int main(){
	cin >> N >> K;
	cin >> s;
	int l = 1, r= N+1;
	while(r-l>1){
		int mid = (l+r)/2;
		if(check(mid))l=mid;
		else r = mid;
	}
	cout << l << endl;
	return 0;
}