Educational Codeforces Round 11——C. Hard Process（YY）

C. Hard Process

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array a with n elements. Each element of a is either 0 or 1.

Let's denote the length of the longest subsegment of consecutive elements in a, consisting of only numbers one, as f(a). You can change no more than k zeroes to ones to maximize f(a).

Input

The first line contains two integers n and k (1 ≤ n ≤ 3·105, 0 ≤ k ≤ n) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 1) — the elements of a.

Output

On the first line print a non-negative integer z — the maximal value of f(a) after no more than k changes of zeroes to ones.

On the second line print n integers aj — the elements of the array a after the changes.

If there are multiple answers, you can print any one of them.

Examples

input

7 1
1 0 0 1 1 0 1

output

4
1 0 0 1 1 1 1

input

10 2
1 0 0 1 0 1 0 1 0 1

output

5
1 0 0 1 1 1 1 1 0 1

记录0出现的位置，随便写几组数据可以发现要判断起始点是否为0的情况。

若起始点为0，则要判断sum[r]-sum[l]+1<=k；否则要判断sum[r]-sum[l]<=k。然后再判断一下k是否为0即可

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int N=300010;
int sufix[N];
int pos[N];
int main (void)
{
	ios::sync_with_stdio(false);
	int n,k,i,j;
	while (cin>>n>>k)
	{
		memset(sufix,0,sizeof(sufix));
		memset(pos,0,sizeof(pos));
		for (i=1; i<=n; i++)
		{
			cin>>pos[i];
			sufix[i]=sufix[i-1]+(pos[i]==0);
		}
		int l,r,dx,al,ar;
		l=1,r=1,dx=0,ar=al=0;
		while (l<=n&&r<=n)
		{
			bool flag=0;
			if(pos[l])
			{
				if(sufix[r]-sufix[l]<=k)
				{
					if(r-l+1>=dx)
					{
						dx=r-l+1;
						al=l;
						ar=r;
					}										
				}
				else
				{
					l++;
				}
				r++;	
			}
			else
			{
				if(sufix[r]-sufix[l]+1<=k)
				{
					if(r-l+1>=dx)
					{
						dx=r-l+1;
						al=l;
						ar=r;
					}										
				}
				else
				{
					l++;
				}
				r++;	
			}
		}
		for (i=al; i<=ar; i++)
		{
			pos[i]=1;
		}
		if(ar==0&&al==0)
			cout<<0<<endl;
		else
			cout<<ar-al+1<<endl;
		for (i=1; i<=n; i++)
		{
			if(i==n)
				cout<<pos[i]<<endl;
			else
				cout<<pos[i]<<" ";
		}
	}
	return 0;
}