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;
using pii = pair<int, int>;

constexpr ll inf = 1e18;

int main() {
    int n; cin >> n;
    vector<ll> a(n);
    for(auto& aa : a) cin >> aa;

    vector<map<ll, int>> dp(n + 1);
    vector<map<ll, pair<ll, int>>> pre(n + 1);
    dp[0][-inf] = 0;
    for(int i = 0; i < n; ++i) {
        ll sum = 0;
        for(int j = i + 1; j <= n; ++j) {
            sum += a[j - 1];
            auto it = dp[i].lower_bound(sum);
            if(it == dp[i].begin()) continue;
            it = prev(it);
            { // update
                auto nit = dp[j].lower_bound(sum);
                while(nit != end(dp[j]) && nit->second <= it->second + 1) {
                    pre[j].erase(nit->first);
                    nit = dp[j].erase(nit);
                }
                if(nit == begin(dp[j]) || prev(nit)->second < it->second + 1) {
                    dp[j][sum] = it->second + 1;
                    pre[j][sum] = {it->first, i};
                }
            }
        }
    }

    int max_len = 0;
    ll cur_sum = 0, cur_idx = n;
    for(auto const& p : dp[n]) {
        if(max_len < p.second) {
            max_len = p.second;
            cur_sum = p.first;
        }
    }
    vector<int> ans;
    while(pre[cur_idx].count(cur_sum)) {
        const auto p = pre[cur_idx][cur_sum];
        ans.push_back(p.second);
        cur_sum = p.first, cur_idx = p.second;
    }
    ans.pop_back();
    reverse(begin(ans), end(ans));

    cout << ans.size() + 1 << endl;
    for(int i = 0; i < (int)ans.size(); ++i) {
        cout << ans[i] << " \n"[i + 1 == (int)ans.size()];
    }
}