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 <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <cstring>
#include <functional>
#include <cmath>
#include <complex>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep1(i,n) for(int i=1;i<=(n);++i)
#define all(c) (c).begin(),(c).end()
#define fs first
#define sc second
#define pb push_back
#define show(x) cout << #x << " " << x << endl
typedef long long ll;
int N;
ll a[4000],sum[4001];
short int inf=10000,p2[4001];
vector<short int> seg[4001];
short int pl[4001][8191];
vector<ll> vals[4001];
int query(int a,int b,int l,int r,int k,int id){
	if(b<=l||r<=a) return -inf;
	if(a<=l&&r<=b) return seg[id][k];
	return max(query(a,b,l,(l+r)/2,2*k+1,id),query(a,b,(l+r)/2,r,2*k+2,id));
}
int main(){
	cin>>N;
	rep(i,N) cin>>a[i];
	rep(i,N+1){
		p2[i]=1;
		while(p2[i]<=i) p2[i]*=2;
	}
	rep(i,N+1) seg[i].resize(p2[i]*2);
	rep(i,N) sum[i+1]=sum[i]+a[i];
	rep(l,N){
		for(int r=l+1;r<=N;r++){
			vals[r].pb(sum[r]-sum[l]);
		}
	}
	rep1(r,N){
		sort(all(vals[r]));
		vals[r].erase(unique(all(vals[r])),vals[r].end());
	}
	for(int l=0;l<N;l++){
		for(int r=l+1;r<=N;r++){
//			printf("(l,r)=%d,%d\n",l,r);
			ll s=sum[r]-sum[l];
			int id=lower_bound(all(vals[l]),s)-vals[l].begin();
			int v=query(0,id,0,p2[l],0,l);
			if(l==0) v=0;
			if(l==0||v>=1){
				int id2=lower_bound(all(vals[r]),s)-vals[r].begin();
				if(v+1>seg[r][p2[r]-1+id2]){
					seg[r][p2[r]-1+id2]=v+1;
					pl[r][p2[r]-1+id2]=l;
					//printf("seg[%d][%d]=%d\n",r,p2-1+id2,seg[r][p2-1+id2]);
					int p=(p2[r]-1+id2-1)/2;
					while(true){
						if(seg[r][p*2+1]>seg[r][p*2+2]){
							seg[r][p]=seg[r][p*2+1],pl[r][p]=pl[r][p*2+1];
						}else{
							seg[r][p]=seg[r][p*2+2],pl[r][p]=pl[r][p*2+2];
						}
						if(p==0) break;
						p=(p-1)/2;
					}
				}
			}
		}
	}
	cout<<seg[N][0]<<endl;
	vector<int> ans;
	int now=N;
	while(true){
		now=pl[now][0];
		if(now==0) break;
		ans.pb(now);
	}
	sort(all(ans));
	rep(i,ans.size()){
		if(i==ans.size()-1) cout<<ans[i]<<endl;
		else cout<<ans[i]<<" ";
	}
}