二分--CodeForces - 660C(二分的应用，也可以用尺取)

                                        A - Hard Process

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 a_i (0 ≤ a_i ≤ 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 kchanges of zeroes to ones.

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

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

Example

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

          

         这道题题意：给定0.1组成的数组，可以改变k个0使其为1，问最终可以得到的连续的1的最大长度。
用一个数组记录当前位一共有多少个0,暴力枚举最长串的最后一位,二分查找最长串的第一个1的位置;更新结果并记录好最长串的开始和结束位置,最后再输出就好啦;

//  main.cpp
//  temp
//
//  Created by Sly on 2017/2/27.
//  Copyright © 2017年 Sly. All rights reserved.
//

#include    //二分
#include 
#include 
#include 
#define N 100000*3
int t[N];
int a[N];
int n,k;
int Bin(int x)   //二分
{
    int mid;
    int l=1;
    int r=x;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(t[x]-t[mid-1]<=k)
            r=mid-1;
        else l=mid+1;
    }
    return l;
}
int main()
{
    int i;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        t[0]=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(!a[i])t[i]=t[i-1]+1;
            else t[i]=t[i-1];
        }
        int ans=0;
        int l=0,r=0;
        for(i=1;i<=n;i++)
        {
            int f=Bin(i);
            if(i-f+1>ans)
            {
                ans=i-f+1;
                l=f;
                r=i;
            }
        }
        printf("%d\n",ans);
        for(i=1;i<=n;i++)
        {
            if(i>=l&&i<=r)
            {
                printf("1");
                if(i!=n)printf(" ");
                else printf("\n");
            }
            else
            {
                printf("%d",a[i]);
                if(i!=n)printf(" ");
                else printf("\n");
            }
        }
    }
    return 0;
}