#include<stdio.h>
#include<vector>
#include<map>
#include<algorithm>
#define INF 1000000
#define LINF 1000000000000LL
using namespace std;
typedef pair<long long,int> P;
long long a[4000];
vector<P> dp[4001];
int r[4000];
vector<int> pos;
int max(int a,int b){return a>b?a:b;}

int main(){
	int n,i,j;
	scanf("%d",&n);
	for(i=0;i<n;i++)scanf("%lld",&a[i]);
	dp[0].push_back(P(-LINF,0));
	for(i=0;i<n;i++){
		int m=dp[i].size();
		sort(dp[i].begin(),dp[i].end());
		r[0]=-INF;
		for(j=1;j<=m;j++)r[j]=max(r[j-1],dp[i][j-1].second);
		long long s=0;
		for(j=i+1;j<=n;j++){
			s+=a[j-1];
			int p=lower_bound(dp[i].begin(),dp[i].end(),P(s,0))-dp[i].begin();
			dp[j].push_back(P(s,r[p]+1));
		}
	}
	P ans=P(0,-INF);
	for(i=0;i<dp[n].size();i++)if(ans.second<dp[n][i].second)ans=dp[n][i];
	printf("%d\n",ans.second);
	int p=n;
	int t=ans.second;
	while(t>0){
		long long s=0;
		while(1){
			p--;
			s+=a[p];
			for(i=0;i<dp[p].size();i++)if(dp[p][i].first<s&&dp[p][i].second==t-1)break;
			if(i<dp[p].size())break;
		}
		if(p>0)pos.push_back(p);
		t--;
	}
	reverse(pos.begin(),pos.end());
	for(i=0;i<pos.size();i++)printf("%d ",pos[i]);
	puts("");
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<stdio.h>
#include<vector>
#include<map>
#include<algorithm>
#define INF 1000000
#define LINF 1000000000000LL
using namespace std;
typedef pair<long long,int> P;
long long a[4000];
vector<P> dp[4001];
int r[4000];
vector<int> pos;
int max(int a,int b){return a>b?a:b;}

int main(){
	int n,i,j;
	scanf("%d",&n);
	for(i=0;i<n;i++)scanf("%lld",&a[i]);
	dp[0].push_back(P(-LINF,0));
	for(i=0;i<n;i++){
		int m=dp[i].size();
		sort(dp[i].begin(),dp[i].end());
		r[0]=-INF;
		for(j=1;j<=m;j++)r[j]=max(r[j-1],dp[i][j-1].second);
		long long s=0;
		for(j=i+1;j<=n;j++){
			s+=a[j-1];
			int p=lower_bound(dp[i].begin(),dp[i].end(),P(s,0))-dp[i].begin();
			dp[j].push_back(P(s,r[p]+1));
		}
	}
	P ans=P(0,-INF);
	for(i=0;i<dp[n].size();i++)if(ans.second<dp[n][i].second)ans=dp[n][i];
	printf("%d\n",ans.second);
	int p=n;
	int t=ans.second;
	while(t>0){
		long long s=0;
		while(1){
			p--;
			s+=a[p];
			for(i=0;i<dp[p].size();i++)if(dp[p][i].first<s&&dp[p][i].second==t-1)break;
			if(i<dp[p].size())break;
		}
		if(p>0)pos.push_back(p);
		t--;
	}
	reverse(pos.begin(),pos.end());
	for(i=0;i<pos.size();i++)printf("%d ",pos[i]);
	puts("");
	return 0;
}
```