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<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);

ll mod_pow(ll x, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x % m;
		x = x * x % m; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n % mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a * a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 17;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

struct SegT {
private:
	int sz; vector<LP> node;
	const LP init_c = { INF,10000 };
public:
	
	SegT(int n) {
		sz = 1;
		while (sz < n)sz *= 2;
		node.resize(2 * sz - 1, init_c);
	}
	LP f(LP a, LP b) {
		return min(a, b);
	}
	void update(int k, LP a) {
		k += sz - 1;
		node[k] = a;
		while (k > 0) {
			k = (k - 1) / 2;
			node[k] = f(node[k * 2 + 1], node[k * 2 + 2]);
		}
	}
	LP query(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0)r = sz;
		if (r <= a || b <= l)return init_c;
		else if (a <= l && r <= b)return node[k];
		else {
			LP vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
			LP vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
			return f(vl, vr);
		}
	}
};

void solve() {
	int n; cin >> n;
	vector<ll> a(n), ra(n + 1);
	rep(i, n) {
		cin >> a[i]; ra[i + 1] = ra[i] + a[i];
	}
	vector<ll> dp(n + 1, INF);
	dp[0] = -INF;
	int ans = 0;

	vector<vector<int>> pres;
	while (true) {
		vector<ll> nex(n + 1);
		SegT st(n);

		vector<LP> v;
		rep(i, n) {
			v.push_back({ dp[i] + ra[i],i });
		}
		sort(all(v));

		vector<int> trans(n);
		rep(i, v.size()) {
			trans[v[i].second] = i;
		}

		vector<int> pre(n + 1);
		rep(i, n + 1) {
			int ri = lower_bound(all(v), LP{ ra[i],-1 }) - v.begin();
			LP p = st.query(0, ri);
			ll val = p.first;
			int fr = p.second;
			pre[i] = fr;
			nex[i] = val + ra[i];
			if (i < n) {
				int id = trans[i];
				st.update(id, { -ra[i],i });
			}
		}
		swap(dp, nex);

		pres.push_back(pre);
		if (dp[n] < INF / 10) {
			ans++;
		}
		else break;
	}
	/*rep(i, pres.size()){
		rep(j, n + 1) {
			cout << pres[i][j] << " ";
		}
		cout << "\n";
	}*/
	vector<int> anss;
	int c1 = pres.size() - 2;
	int c2 = n;
	rep(i, ans - 1) {
		c2 = pres[c1][c2];
		c1--;
		anss.push_back(c2);
	}
	reverse(all(anss));
	cout << ans << "\n";
	rep(i, ans - 1) {
		if (i > 0)cout << " ";
		cout << anss[i];
	}
	cout << "\n";
}




signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	init_f();
	//int t; cin >> t; rep(i, t)
	solve();
	return 0;
}