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 <vector>
#include <cassert>
#include <algorithm>
#include <limits>
using namespace std;
using ll = long long;
template <typename T>
constexpr T INF = numeric_limits<T>::max() / 8;
constexpr int MAXN = 4000;
int N;
ll sum[MAXN];
ll getsum(const int inf, const int sup)  // [inf, sup)
{
    assert(0 <= inf);
    assert(sup <= N);
    assert(inf < sup);
    return sum[sup - 1] - (inf == 0 ? 0LL : sum[inf - 1]);
}
int dp[MAXN + 1][MAXN + 1];  //最後の区間が[i,j)である増加列の項数の最大値 (i < j)
int pre[MAXN + 1][MAXN + 1];
int main()
{
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> sum[i];
    }
    for (int i = 1; i < N; i++) {
        sum[i] += sum[i - 1];
    }
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= N; j++) {
            pre[i][j] = -1;
        }
    }
    for (int j = 1; j <= N; j++) {
        dp[0][j] = 1;
    }
    using P = pair<ll, int>;
    using PP = pair<P, int>;
    for (int mid = 1; mid < N; mid++) {
        vector<PP> v;
        for (int inf = 0; inf < mid; inf++) {
            v.push_back({{getsum(inf, mid), -inf}, dp[inf][mid]});
        }
        for (int sup = mid + 1; sup <= N; sup++) {
            v.push_back({{getsum(mid, sup), -sup}, -1});
        }
        sort(v.begin(), v.end());
        int maxi = -INF<int>;
        int maxind = -1;
        for (const auto& pp : v) {
            const int ind = -pp.first.second;
            const int value = pp.second;
            if (ind > mid) {
                dp[mid][ind] = maxi + 1;
                pre[mid][ind] = maxind;
            } else {
                if (maxi <= value) {
                    maxi = value;
                    maxind = ind;
                }
            }
        }
    }

    int num = -1;
    int left = 0;
    for (int i = 0; i < N; i++) {
        if (num <= dp[i][N]) {
            left = i;
            num = dp[i][N];
        }
    }
    cout << num << endl;
    vector<int> sep;
    int right = N;
    while (left > 0) {
        sep.push_back(left);
        const int L = left;
        left = pre[left][right];
        right = L;
    }
    reverse(sep.begin(), sep.end());
    for (const int l : sep) {
        cout << l << " ";
    }
    cout << endl;

    return 0;
}