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 <algorithm>
#include <vector>
#include <string>
#include <memory.h>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <set>
#include <map>
#include <cctype>
#include <iomanip>
#include <sstream>
#include <cctype>
#include <fstream>
#include <cmath>
using namespace std;

#define rep(i, n) for(int i = 0; i< (int)(n); i++)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define pb(e) push_back(e)
#define foreach(c, i) for(iter(c) i = (c).begin(); i != c.end(); ++i)

typedef long long ll;
typedef pair<ll, ll> P;
const int INF = 1000 * 1000 * 1000 + 7;
const double EPS = 1e-10;

const int sz = 4010;
int dp[sz][sz];
int last[sz][sz];
ll A[sz];
P S[sz];

int main(){
  int n;
  cin >> n;
  rep(i, n) cin >> A[i];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 0;
  rep(i, n) dp[0][i+1] = 1;
  
  for(int l = 1; l < n; l++){
    vector<pair<ll, P> > v;
    ll suml = 0;
    ll sumr = 0;
    for(int i = l - 1; i >= 0; i--){
      suml += A[i];
      v.push_back(make_pair(suml, P(dp[i][l], i)));
    }
    sort(v.begin(), v.end());
    S[0] = P(-INF, -1);
    rep(i, v.size()){
      if(v[i].second.first > S[i].first)  S[i+1] = v[i].second;
      else S[i+1] = S[i];
    }
    for(int r = l + 1; r <= n; r++){
      sumr += A[r-1];
      int ind = lower_bound(v.begin(), v.end(), make_pair(sumr, P(0, 0)))-v.begin();
      dp[l][r] = S[ind].first + 1;
      last[l][r] = S[ind].second;
    }
  }
  
  int v = 0;
  int pre;
  int best = 0;
  
  rep(i, n) if(best < dp[i][n]) {
    best = dp[i][n];
    v = i;
  }
  pre = n;
  vector<int> res;
  rep(i, best - 1){
    res.push_back(v);
    pre = v;
    v = last[v][n];
  }
  reverse(res.begin(), res.end());
  cout << best << endl;
  rep(i, res.size()){
    if(i > 0) cout << ' ';
    cout << res[i];
  }
  cout << endl;
  //rep(i, n + 1)rep(j, i) cout << j << "  " << i << " " << dp[j][i] << endl;
  return 0;
}