#include <bits/stdc++.h>
#define INF 1LL<<60
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;

int n;
ll a[4005];
ll sum[4005];
vector<ll> dp[4005];
vector<int> prevs[4005];
P val[4005];

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%lld",&a[i]);
		sum[i+1]+=sum[i]+a[i];
		val[i]=P(sum[i+1],i);
	}
	sort(val,val+n);
	for(int i=0;i<=n;i++){
		dp[i]=vector<ll>(i+1,INF);
		prevs[i]=vector<int>(i+1,-1);
	}
	dp[0][0]=-INF;
	for(int i=0;i<n;i++){
		stack<int> st;
		int l=0;
		while(l<=i && dp[i][l]==INF){
			l++;
		}
		for(int j=i-1;j>=0;j--){
			dp[i][j]=min(dp[i][j],dp[i][j+1]);
		}
		for(int j=0;j<n;j++){
			int pos=val[j].second;
			if(pos<i)continue;
			ll vp=val[j].first-sum[i];
			while(l<=i && vp>dp[i][l]){
				if(dp[pos+1][l+1]>vp){
					//printf("%d %d-> %d %lld\n",i,l,pos,vp);
					dp[pos+1][l+1]=vp;
					prevs[pos+1][l+1]=i;
				}
				if(l+1<=i && vp>dp[i][l+1]){
					l++;
				}else{
					break;
				}
			}
		}
	}
	for(int i=n;i>=1;i--){
		if(dp[n][i]!=INF){
			printf("%d\n",i);
			vector<int> vec;
			int p=n;
			for(int j=i;j>1;j--){
				p=prevs[p][j];
				vec.push_back(p);
			}
			sort(vec.begin(),vec.end());
			for(int j=0;j<i-1;j++){
				printf("%d%c",vec[j],j+1==(i-1)?'\n':' ');
			}
			return 0;
		}
	}
	return 0;
} 

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <bits/stdc++.h>
#define INF 1LL<<60
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;

int n;
ll a[4005];
ll sum[4005];
vector<ll> dp[4005];
vector<int> prevs[4005];
P val[4005];

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%lld",&a[i]);
		sum[i+1]+=sum[i]+a[i];
		val[i]=P(sum[i+1],i);
	}
	sort(val,val+n);
	for(int i=0;i<=n;i++){
		dp[i]=vector<ll>(i+1,INF);
		prevs[i]=vector<int>(i+1,-1);
	}
	dp[0][0]=-INF;
	for(int i=0;i<n;i++){
		stack<int> st;
		int l=0;
		while(l<=i && dp[i][l]==INF){
			l++;
		}
		for(int j=i-1;j>=0;j--){
			dp[i][j]=min(dp[i][j],dp[i][j+1]);
		}
		for(int j=0;j<n;j++){
			int pos=val[j].second;
			if(pos<i)continue;
			ll vp=val[j].first-sum[i];
			while(l<=i && vp>dp[i][l]){
				if(dp[pos+1][l+1]>vp){
					//printf("%d %d-> %d %lld\n",i,l,pos,vp);
					dp[pos+1][l+1]=vp;
					prevs[pos+1][l+1]=i;
				}
				if(l+1<=i && vp>dp[i][l+1]){
					l++;
				}else{
					break;
				}
			}
		}
	}
	for(int i=n;i>=1;i--){
		if(dp[n][i]!=INF){
			printf("%d\n",i);
			vector<int> vec;
			int p=n;
			for(int j=i;j>1;j--){
				p=prevs[p][j];
				vec.push_back(p);
			}
			sort(vec.begin(),vec.end());
			for(int j=0;j<i-1;j++){
				printf("%d%c",vec[j],j+1==(i-1)?'\n':' ');
			}
			return 0;
		}
	}
	return 0;
} 

```