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>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

const int64_t INF = INT64_C(100000000000000);

int main() {
  int n;
  cin>>n;
  vector<int64_t> a(n);
  REP(i,n)cin>>a[i];
  vector<map<int64_t, pair<int,int>>> dp(n+1);
  dp[0][-INF] = make_pair(0,0);
  REP(i,n){
    int64_t sum = 0;
    for (int j=i;j>=0;--j) {
      sum += a[j];
      auto itr = dp[j].lower_bound(sum);
      if (itr != begin(dp[j])) {
        --itr;
        int l = itr->second.first + 1;
        auto itr2 = dp[i+1].upper_bound(sum);
        while(itr2 != end(dp[i+1]) && itr2->second.first <= l) itr2 = dp[i+1].erase(itr2);
        if (itr2 != begin(dp[i+1])) {
          --itr2;
          if (itr2 == end(dp[i+1]) || l > itr2->second.first)
            dp[i+1][sum] = make_pair(l,j);
        } else {
          dp[i+1][sum] = make_pair(l,j);
        }
      }
    }
  }
  cout << dp.back().rbegin()->second.first << endl;
  vector<int> pos;
  int j = n;
  while(j){
    j = dp[j].rbegin()->second.second;
    pos.push_back(j);
  }
  reverse(begin(pos),end(pos));
  REP(i,pos.size()-1){
    if(i)cout<<' ';
    cout<<(pos[i]+1);
  }
  cout<<endl;
  return 0;
}