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;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  int n;
  cin>>n;
  vector<int> as(n);
  for(int i=0;i<n;i++) cin>>as[i];

  vector<ll> sm(n+1,0);
  for(int i=0;i<n;i++) sm[i+1]=sm[i]+as[i];

  const ll MUL = 1e4;
  const ll OFS = 1e14;
  auto P1=[&](ll x,int y){return (x+OFS)*MUL+y;};
  vector< vector<ll> > G(n);
  for(int l=0;l<n;l++){
    for(int r=l+1;r<=n;r++)
      G[l].emplace_back(P1(sm[r]-sm[l],r));
    sort(G[l].begin(),G[l].end());
  }

  using P2 = pair<short, short>;
  vector< vector<P2> > dp(n+1,vector<P2>(n+1,P2(-n,0)));
  for(int l=0;l<n;l++){
    for(auto pp:G[l]){
      int r=pp%MUL;
      if(l==0) chmax(dp[l][r],P2(1,-1));

      int x=lower_bound(G[l].begin(),G[l].end(),P1(sm[r]-sm[l],r))-G[l].begin();
      if(x+1<(int)G[l].size())
        chmax(dp[l][G[l][x+1]%MUL],dp[l][r]);

      if(r==n) continue;
      int k=upper_bound(G[r].begin(),G[r].end(),P1(sm[r]-sm[l],n))-G[r].begin();
      if(k<(int)G[r].size())
        chmax(dp[r][G[r][k]%MUL],P2(dp[l][r].first+1,l));
    }
  }

  P2 ans(0,-1);
  for(int l=0;l<n;l++)
    chmax(ans,dp[l][n]);

  int l=-1,r=n;
  for(int i=0;i<n;i++)
    if(ans==dp[i][n]) l=i;

  cout<<ans.first<<endl;
  vector<int> res;
  while(l!=0){
    res.emplace_back(l);
    int nl=dp[l][r].second;
    int nr=l;
    l=nl;r=nr;
  }
  reverse(res.begin(),res.end());
  for(int i=0;i<(int)res.size();i++){
    if(i) cout<<" ";
    cout<<res[i];
  }
  cout<<endl;
  return 0;
}