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:
#define _USE_MATH_DEFINES
#include <algorithm>
#include <cstdio>
#include <functional>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<ll, int> ll_i;
typedef pair<double, int> d_i;
typedef pair<ll, ll> ll_ll;
typedef pair<double, double> d_d;
struct edge { int u, v; ll w; };

ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-10;

int main() {
	int n; cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	vector<ll> sum(n + 1);
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i - 1];
	vector<ll_i> p(n + 1);
	for (int i = 0; i <= n; i++) p[i] = ll_i(sum[i], i);
	sort(p.begin(), p.end());
	vector<int> v(n + 1);
	for (int i = 0; i <= n; i++) v[i] = p[i].second;
	vector< vector<int> > dp(n + 1, vector<int>(n + 1));
	vector< vector<int> > prev(n + 1, vector<int>(n + 1, -1));
	for (int i = 1; i <= n; i++) dp[0][i] = 1;
	for (int i = 0; i <= n; i++) {
		int k = n, x = 0, p;
		for (int _k = 0; _k <= n; _k++) {
			int _j = v[_k];
			if (_j <= i) continue;
			while (k >= 0) {
				int j = v[k];
				if (j < i) {
					if (sum[i] - sum[j] >= sum[_j] - sum[i]) break;
					if (dp[j][i] > x) {
						p = j;
						x = dp[j][i];
					}
				}
				k--;
			}
			if (x > 0) {
				dp[i][_j] = x + 1;
				prev[i][_j] = p;
			}
		}
	}
	int i, m = 0;
	for (int j = 0; j < n; j++)
		if (dp[j][n] > m) {
			i = j;
			m = dp[j][n];
		}
	vector<int> c;
	int j = n;
	while (i > 0) {
		c.push_back(i);
		int _i = i;
		i = prev[i][j];
		j = _i;
	}
	reverse(c.begin(), c.end());
	cout << m << endl;
	for (int k = 0; k < m - 1; k++) {
		cout << c[k];
		if (k < m - 2) cout << ' ';
	}
	cout << endl;
}