Problem: Time passed and Taro became a high school student. Under the influence of my older brother who was a college student, I began to be interested in computer science. Taro read through computer science textbooks and learned that there was a famous problem called the "longest increase subsequence problem." Taro understood this problem, but was wondering if he could create a similar problem himself. Therefore, Taro created the following problem as a result of trial and error.

* There is a sequence A consisting of n integers
* Decompose the sequence A into m sequences with m-1 delimiters. In addition, each sequence after decomposition must always include one or more numbers.
* If the m numbers created by adding all the integers in each of these m sequences are arranged in the order of the original sequence, it becomes a strict increasing sequence (that is, the resulting sequence is Bi <Bi + 1). I want to do it.
* The goal is to maximize the length m of the final sequence B.



For example, consider the case where the sequence A is {5, -4,10, -3,8}.
Prepare a sequence C that represents the position of the delimiter, and set C = {2,4}.
At this time, the sequence A is divided into the parts (5, -4), (10, -3), (8), and when these interiors are added together, they become 1,7,8, respectively, and the resulting sequence B is {1, 7,8}.

<image>

Taro named this problem the "longest increase subsequence problem" and also considered an algorithm to solve it.
And I contacted you as a member of society to check it. Your job is to create a program to solve the problems created by Taro who has grown up.
Since it is my job to check, it also outputs the position of the m-1 delimiter. In addition, there may be multiple such delimiters for the optimum m, but if this m is output correctly, one of the possible ones should be output.

Constraints

> 1 ≤ n ≤ 4000
> | Ai | ≤ 108
>

* For the integer k, | k | represents the absolute value of k

Input

N + 1 integers are given, separated by line breaks.

> n
> A1
> A2
> ...
> An
>

* n represents the length of the given sequence A
* Ai represents the i-th element of the sequence A.

Output

> m
> C1 C2 .. Cm-1
>

* The first line outputs one integer m. m represents the length of the final sequence B. The second line outputs m-1 integer Ci separated by blanks. Ci represents the i-th delimiter when the sequence A is appropriately divided into m parts. See the figure for the definition of the delimiter. Note that 1 ≤ Ci <n is satisfied. The m-1 integer sequence C in the second row must be arranged in ascending order. Therefore, if i <j, then Ci <Cj holds.


* If m = 1, the second line will be blank.
* When the sequence B is generated from the sequence A according to the sequence C, when the sequence B is not an increasing sequence (that is, when i (1 ≤ i <m) such that Bi ≥ Bi + 1 exists), it becomes WrongAnswer.

Examples

Input

3
1
2
4


Output

3
1 2


Input

3
2
2
2


Output

2
1


Input

3
4
2
1


Output

1
(空行)

Solution (in CPP):

Solution:
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
int b[4100];
long long sum[4100];
int dp[4100][4100];
int rev[4100][4100];
int ans[4100];
int main(){
	int a;scanf("%d",&a);
	for(int i=0;i<a;i++)scanf("%d",b+i);
	for(int i=0;i<a;i++){
		sum[i+1]=sum[i]+b[i];
	}
	for(int i=0;i<a;i++)dp[0][i+1]=1;
	for(int i=1;i<a;i++){
		vector<pair<long long,int> >L;
		vector<pair<long long,int> >R;
		for(int j=0;j<i;j++)L.push_back(make_pair(sum[i]-sum[j],j));
		for(int j=i+1;j<=a;j++)R.push_back(make_pair(sum[j]-sum[i],j));
		std::sort(L.begin(),L.end());
		std::sort(R.begin(),R.end());
		int ind=0;
		int m=-99999999;
		int at=-1;
		for(int j=0;j<R.size();j++){
			while(ind<L.size()&&L[ind].first<R[j].first){
				if(m<dp[L[ind].second][i]){
					m=dp[L[ind].second][i];
					at=L[ind].second;
				}
				ind++;
			}
			dp[i][R[j].second]=m+1;
			rev[i][R[j].second]=at;
		}
	}
	//for(int i=0;i<a;i++)for(int j=i+1;j<=a;j++)printf("%d %d: %d\n",i,j,dp[i][j]);
	int ret=0;
	int at=0;
	for(int i=0;i<a;i++){
		if(ret<dp[i][a]){
			ret=dp[i][a];
			at=i;
		}
	}
	int last=a;
	printf("%d\n",ret);
	for(int i=0;i<ret;i++){
		ans[ret-1-i]=last;
		int to=rev[at][last];
		last=at;at=to;
	}
	for(int i=0;i<ret-1;i++){
		if(i)printf(" ");
		printf("%d",ans[i]);
	}
	printf("\n");
}