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

using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

struct Data{
  int L,R;//[L,R]
  long long sum;
  Data(int L=0,int R=0,long long sum=0):L(L),R(R),sum(sum){}
  bool operator < (const Data& data)const{
    if( sum != data.sum ) return sum < data.sum;
    return L > data.L;
  }
};


const int IINF = INT_MAX, MAX = 4010;

int n,pre[MAX][MAX],dp[MAX];
ll sum[MAX];

int main(){
  cin >> n;
  vector<ll> A(n);
  rep(i,n){
    cin>>A[i];
    if( i == 0 ) sum[i] = A[i];
    else         sum[i] = sum[i-1] + A[i];
  }

  vector<Data> buf;
  rep(i,n+1)REP(j,i+1,n+1)buf.push_back(Data(i,j,sum[j-1]-((i-1>=0)?sum[i-1]:0LL)));
  sort(buf.begin(),buf.end());
  rep(i,MAX)dp[i] = -IINF;
  rep(i,MAX)rep(j,MAX)pre[i][j] = IINF;

  dp[0] = 0LL;

  int maxcost = -IINF;
  rep(i,buf.size()){
    int cur = buf[i].L, next = buf[i].R;
    //cout << "buf[" << i << "] = " << cur << "," << next << " " << buf[i].sum << endl;
    if( dp[cur] == -IINF ) continue;
    if( dp[next] < dp[cur] + 1 ){
      //cout << "update dp[" << next << "] = " << dp[cur] + 1 << " pre[" << next << "] = " << cur << endl;
      dp[next] = dp[cur] + 1;
      pre[dp[next]][next] = cur;
      maxcost = max(maxcost,dp[next]);
    }
  }


  int cur = n,cnt = IINF;
  vector<int> path;

  for(int i = MAX-1; i >= 0; i-- ) if( pre[i][cur] != IINF ){
      cnt = i;
      break;
    }

  assert( cnt != IINF );
  while( cnt >= 0 && pre[cnt][cur] != IINF ){
    path.push_back(cur);
    cur = pre[cnt][cur];
    cnt--;
  }

  if( !path.empty() )reverse(path.begin(),path.end());
  if( path.size() >= 1)path.erase(path.end()-1);

  cout << path.size() + 1 << endl;
  rep(i,path.size()) {
    if( i ) cout << ' ';
    cout << path[i];
  } cout << endl;

  return 0;
}