#include <stdio.h>
#include <vector>
#include <algorithm>
#include <utility>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define mp make_pair
typedef long long Int;
#define INF (1<<28)

int n, a[4096];
int dp[4096][4096], fr[4096][4096];
Int s[4096];

int main() {
    scanf("%d", &n);
    rep (i, n) scanf("%d", a+i);
    rep (i, n) s[i+1] = s[i] + a[i];
    rep (i, n) dp[0][i+1] = 1;
    rep (i, n) {
        vector<pair<Int, int> > v;
        rep (j, i) if (dp[j][i]) v.push_back(mp(s[i]-s[j], j));
        for (int j = i+1; j <= n; j++) v.push_back(mp(s[j]-s[i], -j));
        sort(v.begin(), v.end());
        int mx = 0, mi = -1;
        rep (k, v.size()) {
            const int j = v[k].second;
            if (j < 0) {
                if (mi != -1) {
                    dp[i][-j] = mx + 1;
                    fr[i][-j] = mi;
                }
            }
            else {
                if (mx < dp[j][i]) {
                    mx = dp[j][i];
                    mi = j;
                }
            }
        }
    }
    int ans = -1, pre = -1;
    rep (i, n) {
        if (ans < dp[i][n]) {
            ans = dp[i][n];
            pre = i;
        }
    }
    printf("%d\n", ans);
    vector<int> ax;
    int cur = n;
    while (pre != 0) {
        ax.push_back(pre);
        const int npre = fr[pre][cur];
        cur = pre;
        pre = npre;
    }
    rep (i, ax.size()) {
        if (i) putchar(' ');
        printf("%d", ax[ax.size()-1-i]);
    }
    printf("\n");
    return 0;
}
### Prompt

Please create a solution in CPP to 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 <stdio.h>
#include <vector>
#include <algorithm>
#include <utility>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define mp make_pair
typedef long long Int;
#define INF (1<<28)

int n, a[4096];
int dp[4096][4096], fr[4096][4096];
Int s[4096];

int main() {
    scanf("%d", &n);
    rep (i, n) scanf("%d", a+i);
    rep (i, n) s[i+1] = s[i] + a[i];
    rep (i, n) dp[0][i+1] = 1;
    rep (i, n) {
        vector<pair<Int, int> > v;
        rep (j, i) if (dp[j][i]) v.push_back(mp(s[i]-s[j], j));
        for (int j = i+1; j <= n; j++) v.push_back(mp(s[j]-s[i], -j));
        sort(v.begin(), v.end());
        int mx = 0, mi = -1;
        rep (k, v.size()) {
            const int j = v[k].second;
            if (j < 0) {
                if (mi != -1) {
                    dp[i][-j] = mx + 1;
                    fr[i][-j] = mi;
                }
            }
            else {
                if (mx < dp[j][i]) {
                    mx = dp[j][i];
                    mi = j;
                }
            }
        }
    }
    int ans = -1, pre = -1;
    rep (i, n) {
        if (ans < dp[i][n]) {
            ans = dp[i][n];
            pre = i;
        }
    }
    printf("%d\n", ans);
    vector<int> ax;
    int cur = n;
    while (pre != 0) {
        ax.push_back(pre);
        const int npre = fr[pre][cur];
        cur = pre;
        pre = npre;
    }
    rep (i, ax.size()) {
        if (i) putchar(' ');
        printf("%d", ax[ax.size()-1-i]);
    }
    printf("\n");
    return 0;
}
```