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;
struct X{
    int id;
    ll val;
    int sc;
    X* prev;
};
bool operator<(X lhs,X rhs){
    return lhs.val==rhs.val ? lhs.sc>rhs.sc : lhs.val<rhs.val;
}
using ll=long long;
const ll INF=1e15;
int main(){
    int n;
    cin>>n;
    vector<ll> a(n);
    for(int i=0;i<n;i++) cin>>a[i];
    vector<ll> sum(n+1);
    partial_sum(a.begin(),a.end(),sum.begin()+1);

    vector<vector<X>> dp(n+1);
    dp[0].push_back(X{0,-INF,0,nullptr});

    for(int i=0;i<n;i++){
        vector<X> xs;
        for(int j=0;j<=i;j++){
            ll rangeSum=sum[i+1]-sum[j];
            X x=X{-1,rangeSum,n,nullptr};
            auto it=lower_bound(dp[j].begin(),dp[j].end(),x);
            if(it==dp[j].begin()) continue;
            it=prev(it);
            xs.push_back(X{i+1,rangeSum,(it->sc)+1,&(*it)});
        }
        sort(xs.begin(),xs.end());
        
        dp[i+1].push_back(xs[0]);
        int presc=xs[0].sc;
        for(int j=1;j<xs.size();j++){
            if(xs[j].sc<=presc) continue;
            presc=xs[j].sc;
            dp[i+1].push_back(xs[j]);
        }
    }
    // cerr<<"DP part finished"<<endl;
    X* x=&(dp[n].back());
    vector<int> res;
    
    while(x!=nullptr){
        res.push_back(x->id);
        x=x->prev;
    }
    res.pop_back();
    reverse(res.begin(),res.end());
    
    cout<<res.size()<<endl;
    for(int i=0;i+1<res.size();i++){
        cout<<res[i]<<(i+2==res.size() ? "" : " ");
    }
    cout<<endl;

    return 0;
}