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 <sstream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;
typedef long long ll;

const int    INF = 1000000000;
const int    MOD = 1000000007;
const double EPS = 1e-8;

int main(){
    int N;
    cin >> N;
    vector<ll> A(N);
    vector<ll> sum(N + 1, 0);
    for(int i = 0; i < N; i++){
        cin >> A[i];
        sum[i + 1] = sum[i] + A[i];
    }

    static ll dp[4005][4005] = {}; // minimam of A[0..i] to js
    static int back[4005][4005] = {};
    for(int i = 0; i <= N + 1; i++)
        for(int j = 0; j <= N + 1; j++)
            dp[i][j] = INF;

    dp[0][0] = -INF;
    for(int i = 0; i < N; i++){
        for(int j = 0; j <= i; j++){
            // a[j], a[j + 1], ... a[i]
            ll s = sum[i + 1] - sum[j];
            int k = lower_bound(dp[j], dp[j] + N + 1, s) - dp[j];
            if(k == 0) continue;
            if(dp[i + 1][k] > s){
                dp[i + 1][k] = s;
                back[i + 1][k] = j - 1;
                //printf("dp[%d][%d] = %lld back[%d][%d] = %d\n", i + 1, k, s, i + 1, k, j - 1);
            }
        }
        for(int k = i; k >= 0; k--) dp[i + 1][k] = min(dp[i + 1][k], dp[i + 1][k + 1]);
        //for(int k = 0; k <= i + 2; k++) cout << dp[i + 1][k] << " "; cout << endl;
    }
    int M = lower_bound(dp[N], dp[N] + N + 1, INF) - dp[N] - 1;
    int curi = N - 1;
    int curM = M;
    vector<int> C;
    while(curM >= 2){
        //printf("curi %d curM %d\n", curi, curM);
        curi = back[curi + 1][curM--];
        C.push_back(curi + 1);
    }
    reverse(C.begin(), C.end());
    assert(C.size() == M - 1);
    cout << M << endl;
    for(int i = 0; i < C.size(); i++){
        cout << C[i];
        if(i + 1 < C.size()) cout << " ";
    }
    cout << endl;
    return 0;
}