Problem:
Score : 600 points

Problem Statement

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
//#include <chrono>
//#include <intrin.h>

using namespace std;

typedef long long ll;

inline void solve() {
	int n;
	cin >> n;
	vector<ll> v(n + 1);
	for (int i = 0; i <= n; i++)
		cin >> v[i];
	ll k = 1;

	if (v[0] > 0 && n > 0) {
		cout << -1;
		return;
	}

	vector<ll> levels(n + 1);

	for (int i = 0; i < n; i++) {
		ll ti = (k - v[i]);
		levels[i] = k;
		k = ti * 2;
	}
	levels[n] = v[n];

	ll otv = v[n];
	ll t = v[n];

	for (int i = n - 1; i >= 0; i--) {
		if ((levels[i] - v[i]) * 2 < t) {
			cout << -1;
			return;
		}
		t = min(levels[i] - v[i], t);
		if (t < 0) {
			cout << -1;
			return;
		}
		t += v[i];
		if (t > levels[i]) {
			cout << -1;
			return;
		}
		otv += t;
	}

	k = 1;

	if (otv < 0) {
		cout << -1;
		return;
	}

	cout << otv;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
#if defined(_DEBUG)
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	int q = 1;
	//cin >> q;
	for (; q > 0; q--) {
		solve();
		cout << '\n';
	}
}