#include<cstdio>
#include<vector>
#include<algorithm>

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

using namespace std;

typedef long long ll;

const int INF=1<<29;

int main(){
	int n; scanf("%d",&n);
	ll a[4000];
	rep(i,n) scanf("%lld",a+i);

	vector<ll> h[4001];
	h[0].push_back(-(1LL<<61));
	rep(i,n){
		ll sum=0;
		for(int j=i;j<n;j++){
			sum+=a[j];
			h[j+1].push_back(sum);
		}
	}
	rep(i,n+1){
		sort(h[i].begin(),h[i].end());
		h[i].erase(unique(h[i].begin(),h[i].end()),h[i].end());
	}

	vector<int> dp[4001];
	vector< pair<int,int> > prev[4001];
	rep(i,n+1){
		dp[i].assign(h[i].size(),-INF);
		prev[i].resize(h[i].size());
	}
	dp[0][0]=0;
	rep(i,n){
		// h[i] の二分探索が正しく動くように, dp[i] を単調増加になるように修正
		int pos[4001]; // pos[j] := ( dp[i][j] の値はもともと dp[i][pos[j]] のもの ) ( 経路復元用 )
		rep(j,dp[i].size()) pos[j]=j;
		rep(j,(int)dp[i].size()-1){
			if(dp[i][j+1]<dp[i][j]){
				dp[i][j+1]=dp[i][j];
				pos[j+1]=pos[j];
			}
		}

		ll sum=0;
		for(int j=i;j<n;j++){
			sum+=a[j];
			int k=lower_bound(h[i].begin(),h[i].end(),sum)-h[i].begin()-1;
			if(k==-1) continue;

			int idx=lower_bound(h[j+1].begin(),h[j+1].end(),sum)-h[j+1].begin();
			if(dp[j+1][idx]<dp[i][k]+1){
				dp[j+1][idx]=dp[i][k]+1;
				prev[j+1][idx]=make_pair(i,pos[k]);
			}
		}
	}

	vector<int> ans;
	int i=n,j=max_element(dp[n].begin(),dp[n].end())-dp[n].begin();
	while(1){
		int tmp_i=prev[i][j].first;
		int tmp_j=prev[i][j].second;
		i=tmp_i;
		j=tmp_j;
		if(i==0) break;
		ans.push_back(i);
	}
	reverse(ans.begin(),ans.end());

	printf("%d\n",ans.size()+1);
	rep(i,ans.size()) printf("%d%s",ans[i],i+1<ans.size()?" ":"");
	puts("");

	return 0;
}
### Prompt

Your task is to create 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<vector>
#include<algorithm>

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

using namespace std;

typedef long long ll;

const int INF=1<<29;

int main(){
	int n; scanf("%d",&n);
	ll a[4000];
	rep(i,n) scanf("%lld",a+i);

	vector<ll> h[4001];
	h[0].push_back(-(1LL<<61));
	rep(i,n){
		ll sum=0;
		for(int j=i;j<n;j++){
			sum+=a[j];
			h[j+1].push_back(sum);
		}
	}
	rep(i,n+1){
		sort(h[i].begin(),h[i].end());
		h[i].erase(unique(h[i].begin(),h[i].end()),h[i].end());
	}

	vector<int> dp[4001];
	vector< pair<int,int> > prev[4001];
	rep(i,n+1){
		dp[i].assign(h[i].size(),-INF);
		prev[i].resize(h[i].size());
	}
	dp[0][0]=0;
	rep(i,n){
		// h[i] の二分探索が正しく動くように, dp[i] を単調増加になるように修正
		int pos[4001]; // pos[j] := ( dp[i][j] の値はもともと dp[i][pos[j]] のもの ) ( 経路復元用 )
		rep(j,dp[i].size()) pos[j]=j;
		rep(j,(int)dp[i].size()-1){
			if(dp[i][j+1]<dp[i][j]){
				dp[i][j+1]=dp[i][j];
				pos[j+1]=pos[j];
			}
		}

		ll sum=0;
		for(int j=i;j<n;j++){
			sum+=a[j];
			int k=lower_bound(h[i].begin(),h[i].end(),sum)-h[i].begin()-1;
			if(k==-1) continue;

			int idx=lower_bound(h[j+1].begin(),h[j+1].end(),sum)-h[j+1].begin();
			if(dp[j+1][idx]<dp[i][k]+1){
				dp[j+1][idx]=dp[i][k]+1;
				prev[j+1][idx]=make_pair(i,pos[k]);
			}
		}
	}

	vector<int> ans;
	int i=n,j=max_element(dp[n].begin(),dp[n].end())-dp[n].begin();
	while(1){
		int tmp_i=prev[i][j].first;
		int tmp_j=prev[i][j].second;
		i=tmp_i;
		j=tmp_j;
		if(i==0) break;
		ans.push_back(i);
	}
	reverse(ans.begin(),ans.end());

	printf("%d\n",ans.size()+1);
	rep(i,ans.size()) printf("%d%s",ans[i],i+1<ans.size()?" ":"");
	puts("");

	return 0;
}
```