Paper Work--codeForces 250A

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为避免老板因连续三天以上的公司亏损报告而发怒，需将包含最多两天亏损记录的连续日期分组进最少数量的文件夹中。通过贪心算法实现这一目标，确保每个文件夹内的亏损天数不超过两天。

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Description

Polycarpus has been working in the analytic department of the "F.R.A.U.D." company for as much as n days. Right now his task is to make a series of reports about the company's performance for the last n days. We know that the main information in a day report is value a_i, the company's profit on the i-th day. If a_i is negative, then the company suffered losses on the i-th day.

Polycarpus should sort the daily reports into folders. Each folder should include data on the company's performance for several consecutive days. Of course, the information on each of the n days should be exactly in one folder. Thus, Polycarpus puts information on the first few days in the first folder. The information on the several following days goes to the second folder, and so on.

It is known that the boss reads one daily report folder per day. If one folder has three or more reports for the days in which the company suffered losses (a_i < 0), he loses his temper and his wrath is terrible.

Therefore, Polycarpus wants to prepare the folders so that none of them contains information on three or more days with the loss, and the number of folders is minimal.

Write a program that, given sequence a_i, will print the minimum number of folders.

Input

The first line contains integer n (1 ≤ n ≤ 100), n is the number of days. The second line contains a sequence of integers a₁, a₂, ..., a_n(|a_i| ≤ 100), where a_i means the company profit on the i-th day. It is possible that the company has no days with the negative a_i.

Output

Print an integer k — the required minimum number of folders. In the second line print a sequence of integers b₁, b₂, ..., b_k, where b_j is the number of day reports in the j-th folder.

If there are multiple ways to sort the reports into k days, print any of them.

Sample Input

Input

11
1 2 3 -4 -5 -6 5 -5 -6 -7 6

Output

3
5 3 3

Input

5
0 -1 100 -1 0

Output

1
5

//此题的一个贪心思想是，每个文件夹放2个损失的（如果有）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int A[108];
int B[108];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF&&n)
	{
		memset(B,0,sizeof(B));
		int k=0;
		for(int i=1;i<=n;i++)
		{
			cin>>A[i];
			if(A[i]<0)
			{
				B[++k]=i;
			}
		}
		if(k<=2)
		{
			cout<<1<<endl;
			cout<<n<<endl;
		}
		else//也就是k>2
		{	
			int sum;
			if(k%2==0)
			{
				sum=k/2;
				
			}
			else sum=k/2+1;
			cout<<sum<<endl;
			cout<<B[2]<<" ";
			int weiba=B[2];
			int j=4;
			int t=1;
			while(B[j]!=0&&j!=k)
			{
				t++;
				cout<<B[j]-weiba;
				weiba=B[j];
				j+=2;
				if(t==sum)
				{
					cout<<endl;
				}
				else cout<<" ";
			}
			if(t!=sum)
			{
				cout<<n-weiba<<endl;
			}
		}
	}
	return 0;
}