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 <bits/stdc++.h>
using namespace std;
using ll = long long;

const ll minf = -(ll)1e18;

int main()
{
	int n;
	cin >> n;
	vector<ll> A(n), sum(n + 1);
	for (int i = 0; i < n; i++) {
		cin >> A[i];
		sum[i + 1] = sum[i] + A[i];
	}
	vector<map<ll, int>> v(n + 1);
	vector<int> pre(n + 1);
	v[0][minf] = 0;
	for (int i = 0; i < n; i++) {
		map<int, ll> tmp;
		int tt = 0, ts = 0;
		for (int j = i; j >= 0; j--) {
			auto it = v[j].lower_bound(sum[i + 1] - sum[j]);
			if (it != v[j].begin()) {
				--it;
				if (it->second + 1 > tt || (it->second + 1 == tt && sum[i + 1] - sum[j] < ts)) {
					tt = it->second + 1;
					ts = sum[i + 1] - sum[j];
					pre[i + 1] = j;
				}
				if (tmp.count(it->second + 1)) {
					tmp[it->second + 1] = min(tmp[it->second + 1], sum[i + 1] - sum[j]);
				}
				else {
					tmp[it->second + 1] = sum[i + 1] - sum[j];
				}
			}
		}
		vector<pair<ll, int>> hoge;
		for (auto p : tmp) {
			hoge.emplace_back(p.second, p.first);
		}
		sort(hoge.begin(), hoge.end());
		ll ma = minf;
		for (auto p : hoge) {
			if (p.second > ma) {
				ma = p.second;
				v[i + 1][p.first] = p.second;
			}
		}
	}
	int res = 0;
	for (auto p : v[n]) {
		res = max(res, p.second);
	}
	cout << res << endl;
	int it = n;
	vector<int> vs;
	while (pre[it] != 0) {
		vs.push_back(pre[it]);
		it = pre[it];
	}
	reverse(vs.begin(), vs.end());
	for (int i = 0; i < (int)vs.size(); i++) {
		if (i != 0) cout << ' ';
		cout << vs[i];
	}
	cout << endl;
	return 0;
}