#include <cstdio>
#include <algorithm>
using namespace std;
int n, c, e, ans[4009], pre[4009], dp[4009], ps[4009][4009], l[8002009], r[8002009], p[8002009]; long long s[4009];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", &s[i]), s[i] += s[i - 1];
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j <= n; j++) {
			l[c] = i, r[c] = j, p[c] = c, c++;
		}
	}
	sort(p, p + c, [](int i, int j) { return s[r[i]] - s[l[i]] != s[r[j]] - s[l[j]] ? s[r[i]] - s[l[i]] < s[r[j]] - s[l[j]] : i > j; });
	fill(dp + 1, dp + n + 1, -(1 << 30));
	for (int i = 0; i < c; i++) {
		if (dp[r[p[i]]] < dp[l[p[i]]] + 1) {
			dp[r[p[i]]] = dp[l[p[i]]] + 1;
			ps[r[p[i]]][l[p[i]]] = pre[l[p[i]]];
			pre[r[p[i]]] = l[p[i]];
		}
	}
	printf("%d\n", dp[n]);
	int p1 = n, p2 = pre[n];
	while (p2 != 0) {
		ans[e++] = p2;
		swap(p1, p2);
		p2 = ps[p2][p1];
	}
	for (int i = e - 1; i >= 0; i--) printf(i == 0 ? "%d\n" : "%d ", ans[i]);
	return 0;
}

### Prompt

Please provide a cpp coded solution 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 <cstdio>
#include <algorithm>
using namespace std;
int n, c, e, ans[4009], pre[4009], dp[4009], ps[4009][4009], l[8002009], r[8002009], p[8002009]; long long s[4009];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", &s[i]), s[i] += s[i - 1];
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j <= n; j++) {
			l[c] = i, r[c] = j, p[c] = c, c++;
		}
	}
	sort(p, p + c, [](int i, int j) { return s[r[i]] - s[l[i]] != s[r[j]] - s[l[j]] ? s[r[i]] - s[l[i]] < s[r[j]] - s[l[j]] : i > j; });
	fill(dp + 1, dp + n + 1, -(1 << 30));
	for (int i = 0; i < c; i++) {
		if (dp[r[p[i]]] < dp[l[p[i]]] + 1) {
			dp[r[p[i]]] = dp[l[p[i]]] + 1;
			ps[r[p[i]]][l[p[i]]] = pre[l[p[i]]];
			pre[r[p[i]]] = l[p[i]];
		}
	}
	printf("%d\n", dp[n]);
	int p1 = n, p2 = pre[n];
	while (p2 != 0) {
		ans[e++] = p2;
		swap(p1, p2);
		p2 = ps[p2][p1];
	}
	for (int i = e - 1; i >= 0; i--) printf(i == 0 ? "%d\n" : "%d ", ans[i]);
	return 0;
}

```