#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
using namespace std;

typedef long long LL;

int main(){
	const short SLIM = 20000;
	const LL LLIM = 1LL << 60;

	ios::sync_with_stdio(false);

	int n;
	cin >> n;
	vector<LL> psum(n + 1);
	for(int i = 1; i <= n; ++i){
		LL a;
		cin >> a;
		psum[i] = psum[i - 1] + a;
	}

	vector<vector<pair<LL,short>>> vals(n + 1);
	vector<vector<tuple<short,short,short>>> dp(n + 1);
	for(int i = 0; i <= n; ++i){
		auto &v = vals[i];
		v.reserve(n - i + 1);
		v.emplace_back(LLIM, -1);
		for(int j = i + 1; j <= n; ++j){
			v.emplace_back(psum[j] - psum[i], j);
		}
		sort(v.begin(), v.end());
		dp[i].assign(v.size(), make_tuple(-SLIM, -1, -1));
	}
	dp[0][0] = make_tuple(0, -1, -1);
	for(int i = 0; i <= n; ++i){
		int sz = vals[i].size();
		for(int j = 0; j < sz; ++j){
			if(j > 0){ dp[i][j] = max(dp[i][j], dp[i][j - 1]); }
			if(get<0>(dp[i][j]) < 0){ continue; }
			int k = vals[i][j].second;
			if(k >= 0){
				int l = upper_bound(vals[k].begin(), vals[k].end(), make_pair(vals[i][j].first, SLIM)) - vals[k].begin();
				tuple<short,short,short> tp(get<0>(dp[i][j]) + 1, i, j);
				dp[k][l] = max(dp[k][l], tp);
			}
		}
	}
	
	vector<int> ans;
	int pos = n, h = dp[n].size() - 1;
	while(true){
		auto t = dp[pos][h];
		pos = get<1>(t);
		h = get<2>(t);
		if(pos <= 0){ break; }
		ans.push_back(pos);
	}
	reverse(ans.begin(), ans.end());
	cout << (ans.size() + 1) << '\n';
	bool fst = true;
	for(int x : ans){
		if(!fst){ cout << ' '; }
		cout << x;
		fst = false;
	}
	cout << '\n';
}


### Prompt

Develop a solution in CPP 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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
using namespace std;

typedef long long LL;

int main(){
	const short SLIM = 20000;
	const LL LLIM = 1LL << 60;

	ios::sync_with_stdio(false);

	int n;
	cin >> n;
	vector<LL> psum(n + 1);
	for(int i = 1; i <= n; ++i){
		LL a;
		cin >> a;
		psum[i] = psum[i - 1] + a;
	}

	vector<vector<pair<LL,short>>> vals(n + 1);
	vector<vector<tuple<short,short,short>>> dp(n + 1);
	for(int i = 0; i <= n; ++i){
		auto &v = vals[i];
		v.reserve(n - i + 1);
		v.emplace_back(LLIM, -1);
		for(int j = i + 1; j <= n; ++j){
			v.emplace_back(psum[j] - psum[i], j);
		}
		sort(v.begin(), v.end());
		dp[i].assign(v.size(), make_tuple(-SLIM, -1, -1));
	}
	dp[0][0] = make_tuple(0, -1, -1);
	for(int i = 0; i <= n; ++i){
		int sz = vals[i].size();
		for(int j = 0; j < sz; ++j){
			if(j > 0){ dp[i][j] = max(dp[i][j], dp[i][j - 1]); }
			if(get<0>(dp[i][j]) < 0){ continue; }
			int k = vals[i][j].second;
			if(k >= 0){
				int l = upper_bound(vals[k].begin(), vals[k].end(), make_pair(vals[i][j].first, SLIM)) - vals[k].begin();
				tuple<short,short,short> tp(get<0>(dp[i][j]) + 1, i, j);
				dp[k][l] = max(dp[k][l], tp);
			}
		}
	}
	
	vector<int> ans;
	int pos = n, h = dp[n].size() - 1;
	while(true){
		auto t = dp[pos][h];
		pos = get<1>(t);
		h = get<2>(t);
		if(pos <= 0){ break; }
		ans.push_back(pos);
	}
	reverse(ans.begin(), ans.end());
	cout << (ans.size() + 1) << '\n';
	bool fst = true;
	for(int x : ans){
		if(!fst){ cout << ' '; }
		cout << x;
		fst = false;
	}
	cout << '\n';
}


```