#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>

using namespace std;
typedef long long ll;
const ll INF = 0x7fffffffffll;

struct Entry {
	signed long long sum:40;
	unsigned int     len:12;
	unsigned int     prev:12;
	Entry(ll s = 0, short l = 0, short p = 0) : sum(s), len(l), prev(p) { }
	bool operator<(const Entry &e) const {
		if(sum < e.sum){ return true; }
		if(sum > e.sum){ return false; }
		if(len < e.len){ return true; }
		if(len > e.len){ return false; }
		return prev < e.prev;
	}
};

Entry dp[4001][4002];
short dp_count[4001] = { 0 };
int a[4000];

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	for(int i = 0; i < n; ++i){ cin >> a[i]; }
	dp[0][dp_count[0]++] = Entry(-INF, 0, 0);
	for(int i = 0; i < n; ++i){
		sort(&(dp[i][0]), &(dp[i][dp_count[i]]));
		for(int j = 1; j < dp_count[i]; ++j){
			if(dp[i][j - 1].len >= dp[i][j].len){ dp[i][j] = dp[i][j - 1]; }
		}
		ll s = 0;
		for(int j = i + 1; j <= n; ++j){
			s += a[j - 1];
			Entry *it = lower_bound(
				&(dp[i][0]), &(dp[i][dp_count[i]]), Entry(s, 0, 0));
			if(it == &(dp[i][0])){ continue; }
			dp[j][dp_count[j]++] = Entry(s, (it - 1)->len + 1, j - i);
		}
	}
	Entry maximum = Entry(0, 0, 0);
	for(int i = 0; i < dp_count[n]; ++i){
		if(dp[n][i].len > maximum.len){ maximum = dp[n][i]; }
	}
	int anum = 0, p = n;
	while(maximum.prev > 0){
		p = p - maximum.prev;
		a[anum++] = p;
		for(int i = 0; i < dp_count[p]; ++i){
			if(dp[p][i].sum < maximum.sum && dp[p][i].len == maximum.len - 1){
				maximum = dp[p][i];
				break;
			}
		}
	}
	sort(a, a + anum);
	cout << anum << endl;
	for(int i = 1; i < anum; ++i){
		cout << a[i];
		if(i + 1 != anum){ cout << " "; }
	}
	cout << endl;
	return 0;
}
### Prompt

Create a solution in CPP for 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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>

using namespace std;
typedef long long ll;
const ll INF = 0x7fffffffffll;

struct Entry {
	signed long long sum:40;
	unsigned int     len:12;
	unsigned int     prev:12;
	Entry(ll s = 0, short l = 0, short p = 0) : sum(s), len(l), prev(p) { }
	bool operator<(const Entry &e) const {
		if(sum < e.sum){ return true; }
		if(sum > e.sum){ return false; }
		if(len < e.len){ return true; }
		if(len > e.len){ return false; }
		return prev < e.prev;
	}
};

Entry dp[4001][4002];
short dp_count[4001] = { 0 };
int a[4000];

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	for(int i = 0; i < n; ++i){ cin >> a[i]; }
	dp[0][dp_count[0]++] = Entry(-INF, 0, 0);
	for(int i = 0; i < n; ++i){
		sort(&(dp[i][0]), &(dp[i][dp_count[i]]));
		for(int j = 1; j < dp_count[i]; ++j){
			if(dp[i][j - 1].len >= dp[i][j].len){ dp[i][j] = dp[i][j - 1]; }
		}
		ll s = 0;
		for(int j = i + 1; j <= n; ++j){
			s += a[j - 1];
			Entry *it = lower_bound(
				&(dp[i][0]), &(dp[i][dp_count[i]]), Entry(s, 0, 0));
			if(it == &(dp[i][0])){ continue; }
			dp[j][dp_count[j]++] = Entry(s, (it - 1)->len + 1, j - i);
		}
	}
	Entry maximum = Entry(0, 0, 0);
	for(int i = 0; i < dp_count[n]; ++i){
		if(dp[n][i].len > maximum.len){ maximum = dp[n][i]; }
	}
	int anum = 0, p = n;
	while(maximum.prev > 0){
		p = p - maximum.prev;
		a[anum++] = p;
		for(int i = 0; i < dp_count[p]; ++i){
			if(dp[p][i].sum < maximum.sum && dp[p][i].len == maximum.len - 1){
				maximum = dp[p][i];
				break;
			}
		}
	}
	sort(a, a + anum);
	cout << anum << endl;
	for(int i = 1; i < anum; ++i){
		cout << a[i];
		if(i + 1 != anum){ cout << " "; }
	}
	cout << endl;
	return 0;
}
```