#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#define INF 1e16
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n;
	cin>>n;
	ll a[4001], s[4001];
	s[0]=0;
	for(int i=0; i<n; i++){
		cin>>a[i];
		s[i+1]=s[i]+a[i];
	}
	vector<ll> dp[4001];
	vector<P> prev[4001];
	for(int i=0; i<=n; i++){
		for(int j=0; j<=i; j++){
			dp[i].push_back(INF);
          prev[i].push_back(P(0, 0));
		}
	}
	dp[0][0]=-INF;
	for(int i=1; i<=n; i++){
		int v[4001];
		fill(v, v+i+1, -1);
		for(int j=0; j<i; j++){
			int m1=lower_bound(dp[j].begin(), dp[j].end(), s[i]-s[j])-dp[j].begin();
			if(v[m1]==-1 || s[i]-s[v[m1]]>s[i]-s[j]) v[m1]=j;
		}
		for(int j=i; j>=1; j--){
          if(v[j]==-1) continue;
			if(dp[i][j]>s[i]-s[v[j]]){
				prev[i][j]=P(v[j], j-1);
				dp[i][j]=s[i]-s[v[j]];
			}
		}
      for(int j=i-1; j>=0; j--){
        dp[i][j]=min(dp[i][j], dp[i][j+1]);
      }
	}
	for(int i=n; i>=1; i--){
		if(dp[n][i]<INF){
			cout<<i<<endl;
			P p=prev[n][i];
          vector<int> v;
			while(p.first>0){
				v.push_back(p.first);
				p=prev[p.first][p.second];
			}
          if(v.empty()){
            cout<<endl;
            break;
          }
          for(int j=v.size()-1; j>=0; j--){
            cout<<v[j];
            if(j==0) break;
            cout<<" ";
          }
          cout<<endl;
          break;
		}
	}
    return 0;
}

### Prompt

Generate a CPP solution to the following 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)

### Response

```cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#define INF 1e16
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n;
	cin>>n;
	ll a[4001], s[4001];
	s[0]=0;
	for(int i=0; i<n; i++){
		cin>>a[i];
		s[i+1]=s[i]+a[i];
	}
	vector<ll> dp[4001];
	vector<P> prev[4001];
	for(int i=0; i<=n; i++){
		for(int j=0; j<=i; j++){
			dp[i].push_back(INF);
          prev[i].push_back(P(0, 0));
		}
	}
	dp[0][0]=-INF;
	for(int i=1; i<=n; i++){
		int v[4001];
		fill(v, v+i+1, -1);
		for(int j=0; j<i; j++){
			int m1=lower_bound(dp[j].begin(), dp[j].end(), s[i]-s[j])-dp[j].begin();
			if(v[m1]==-1 || s[i]-s[v[m1]]>s[i]-s[j]) v[m1]=j;
		}
		for(int j=i; j>=1; j--){
          if(v[j]==-1) continue;
			if(dp[i][j]>s[i]-s[v[j]]){
				prev[i][j]=P(v[j], j-1);
				dp[i][j]=s[i]-s[v[j]];
			}
		}
      for(int j=i-1; j>=0; j--){
        dp[i][j]=min(dp[i][j], dp[i][j+1]);
      }
	}
	for(int i=n; i>=1; i--){
		if(dp[n][i]<INF){
			cout<<i<<endl;
			P p=prev[n][i];
          vector<int> v;
			while(p.first>0){
				v.push_back(p.first);
				p=prev[p.first][p.second];
			}
          if(v.empty()){
            cout<<endl;
            break;
          }
          for(int j=v.size()-1; j>=0; j--){
            cout<<v[j];
            if(j==0) break;
            cout<<" ";
          }
          cout<<endl;
          break;
		}
	}
    return 0;
}

```