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<algorithm>
#include<vector>
#include<queue>
#include<cstdio>
using namespace std;

typedef long long ll;
typedef pair<ll,int> P;

#define rep(i,x) for(int i = 0 ; i < x ; i ++)
#define rep1(i,x) for(int i = 1 ; i <= x ; i ++)
#define pb push_back
#define sor(vec) sort(vec.begin(),vec.end())
#define lb(vec,a) lower_bound(vec.begin(),vec.end(),a)
#define ub(vec,a) upper_bound(vec.begin(),vec.end(),a)
#define fr first
#define sc second

const ll INF = 1000000000000000000;

struct bit{
	int siz;
	P s[8200];
	void init(){
		siz = 1<<13;
		rep1(i,siz){
			s[i] = P(-INF,0);
		}
	}
	void updata(int a,P x){
		while(a <= siz){
			if(s[a].fr < x.fr)s[a] = x;
			a += a&-a;
		}
	}
	P query(int a){
		P ret = P(-INF,0);
		if(a == 0)return ret;
		while(a > 0){
			if(ret.fr < s[a].fr)ret = s[a];
			a -= a&-a;
		}
		return ret;
	}
}bit;

int main(){
	int n;
	ll a[4002];
	scanf("%d",&n);
	rep1(i,n)scanf("%lld",&a[i]);
	
	ll sum[4002];
	static P dp[4002][4002];
	sum[0] = 0;
	rep1(i,n)sum[i] = sum[i-1] + a[i];
	rep1(i,n)dp[i][1] = P(sum[i],0);
	for(int j = 1 ; j <= n ; j ++){
		vector<ll> Z;
		for(int i = 1 ; i <= n ; i ++){
			Z.pb(sum[i]+dp[i][j].fr);
		}
		sor(Z);
		bit.init();
		for(int i = j ; i <= n ; i ++){
			dp[i][j+1] = bit.query(lb(Z,sum[i])-Z.begin());
			if(dp[i][j+1].fr == -INF)dp[i][j+1].fr = INF;
			else dp[i][j+1].fr = sum[i]-dp[i][j+1].fr;
			bit.updata((lb(Z,sum[i]+dp[i][j].fr)-Z.begin())+1,P(sum[i],i));
		}
	}
	for(int j = n ; j >= 1 ; j --){
		if(dp[n][j].fr != INF){
			vector<int> ans;
			int loc = n;
			while(1){
				loc = dp[loc][j].sc;
				j --;
				if(loc == 0)break;
				ans.pb(loc);
			}
			printf("%d\n",(int)ans.size()+1);
			reverse(ans.begin(),ans.end());
			rep(i,ans.size()){
				printf("%d%c",ans[i],(i==(int)ans.size()-1)?'\n':' ');
			}
			if(ans.size() == 0)puts("");
			return 0;
		}
	}
}