C. 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 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 11 0 0 1 1 0 1

output

41 0 0 1 1 1 1

input

10 21 0 0 1 0 1 0 1 0 1

output

51 0 0 1 1 1 1 1 0 1

题目意思是改变最少的0，把它变成1，得到最长的全是1的序列。

题目的数据规模明显需要nlogn的算法，直接暴力n^2肯定会超时。

思路：首先定义一个数组sum保存到第i个结点时有多少个1.

然后遍历，从第i个结点开始能得到的最大长度，它等于1的个数加上可以改变成1的0的个数。

#include <cstdio>#include <cmath>#include <algorithm>using namespace std;int a[1000001], sum[1000001];int main( ){        int n, k;scanf("%d %d", &n, &k);for (int i = 1; i <= n; i ++)scanf("%d", &a[i]), sum[i] = sum[i - 1] + a[i];int ans = 0,t1,t2;for(int i=1;i<=n;i++){        int l=i,r=n;        while(l<=r){            int mid=(l+r)>>1;            if(mid-i+1<=k+sum[mid]-sum[i-1])                l=mid+1;            else                r=mid-1;        }        int pos=r;        if(pos-i+1>ans){            ans=pos-i+1;            t1=i;            t2=pos;        }}printf("%d\n", ans);        for (int i = t1; i <= t2; i ++) a[i] = 1;        for (int i = 1; i <= n; i ++)           printf("%d ", a[i]);        printf("\n");return 0;}