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;
typedef long long ll;
typedef pair<ll,ll> P;

map<ll,ll> dp[4001];
map<ll,P> p[4001];

int main() {
  int n;
  cin >> n;
  ll a[n];
  for(int i=0; i<n; i++) cin >> a[i];
  dp[0][-(1LL<<50)]=0;
  p[0][-(1LL<<50)]=P(-1,-1);
  ll d=0;
  for(int i=0; i<n; i++) {
    d+=a[i];
    vector<P> v;
    ll c=0;
    for(int j=i; j>=0; j--) {
      c+=a[j];
      v.push_back(P(c,j));
    }
    sort(v.begin(),v.end());
    ll z=-(1LL<<60);
    for(int j=0; j<v.size(); j++) {
      ll M=-1,x;
      for(map<ll,ll>::iterator it=dp[v[j].second].begin(); it!=dp[v[j].second].end(); it++) {
        if(it->first>=v[j].first) break;
        if(z<it->second) {
          if(M<it->second) {
            M=it->second;
            x=it->first;
          }
        }
      }
      if(M==-1) continue;
      if(!dp[i+1].count(v[j].first)||dp[i+1][v[j].first]<M+1) {
        dp[i+1][v[j].first]=M+1;
        p[i+1][v[j].first]=P(v[j].second,x);
      }
      z=M;
    }
  }
  ll ans=-1;
  P q;
  for(map<ll,ll>::iterator it=dp[n].begin(); it!=dp[n].end(); it++) {
    if(ans<it->second) {
      ans=it->second;
      q=p[n][it->first];
    }
  }
  vector<int> v;
  while(q!=P(-1,-1)) {
    v.push_back(q.first);
    q=p[q.first][q.second];
  }
  reverse(v.begin(),v.end());
  cout << ans << endl;
  for(int i=1; i<v.size(); i++) {
    if(i!=1) cout << " ";
    cout << v[i];
  }
  cout << endl;
  return 0;
}