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 NDEBUG

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <cassert>
#include <numeric>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef tuple<ll, int, int> T;

const int MS = 12;
const int MN = 1<<MS;
ll a[MN], sum[MN];
vector<T> v;
P dp[MN];
int main() {
    int n;
    cin >> n;
    sum[0] = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        sum[i+1] = sum[i]+a[i];
    }
    for (int i = 0; i < n; i++) {
        for (int j = i+1; j <= n; j++) {
            v.push_back(T(sum[j]-sum[i], -i, -j));
        }
    }
    sort(v.begin(), v.end());
    vector<P> res(v.size());
    fill_n(dp, MN, P(-10000, -1));
    dp[0] = P(0, -1);
    for (int i = 0; i < v.size(); i++) {
        ll dummy; int a, b;
        tie(dummy, a, b) = v[i];
        a *= -1; b *= -1;
        res[i] = dp[a];
        if (dp[b].first >= res[i].first+1) continue;
        dp[b] = P(res[i].first+1, i);
    }
    P p = dp[n];
    printf("%d\n", p.first);
    vector<int> rv;
    while (p.second != -1) {
        rv.push_back(-get<2>(v[p.second]));
        p = res[p.second];
    }
    reverse(rv.begin(), rv.end());
    rv.pop_back();
    for (int d: rv) {
        printf("%d ", d);
    } printf("\n");
    return 0;
}