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<iostream>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;

typedef long long Int;
typedef pair<int, int> P;

#define val(x) (a[x.second] - a[x.first - 1])

Int a[4320];
Int p = 0;

vector<int> ans;
vector<P> nums;
int maxdp[4004];
int dp[4004][4004];
int n;
bool comp(const P &x, const P &y){
	if(val(x) == val(y)){
		return x > y;
	}
	return val(x) < val(y);
}

int main(){
	cin >> n;
	for(int i = 1;i <= n;i++)cin >> a[i];
	for(int i = 1;i <= n;i++)a[i] += a[i-1];
	for(int i = 1;i <= n;i++){
		for(int j = i;j <= n;j++){
			nums.push_back(P(i, j));
		}
	}
	sort(nums.begin(), nums.end(), comp);
	for(int i = 0;i < nums.size();i++){
		int from = nums[i].first;
		int to = nums[i].second;
		if(from == 1){
			dp[from][to] = 1;
		}
		else if(maxdp[from-1]){
			dp[from][to] = max(maxdp[from-1] + 1, dp[from][to]);
		}
		maxdp[to] = max(maxdp[to], dp[from][to]);
	}
	int m = maxdp[n];
	cout << m << endl;
	for(int x = n;x >= 1;){
		for(int j = x;j >= 0;j--){
			if(dp[j][x] == m){
				m--;
				ans.push_back(j);
				x = j - 1;
				break;
			}
		}
	}	
	for(int i = ans.size() - 2;i >= 0;i--){
		cout << ans[i]-1;
		if(i)cout << " ";
	}       cout << endl;
	return 0;
}