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: Accepted)

#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
using namespace std;
#define endl "\n"
#define ll long long
#define sz(s) (int)(s.size())
#define INF 0x3f3f3f3f3f3f3f3fLL
#define all(v) v.begin(),v.end()
#define watch(x) cout<<(#x)<<" = "<<x<<endl
const int dr[] { -1, -1, 0, 1, 1, 1, 0, -1 };
const int dc[] { 0, 1, 1, 1, 0, -1, -1, -1 };
#if __cplusplus >= 201402L
template<typename T>
vector<T> create(size_t n) {
	return vector<T>(n);
}
template<typename T, typename ... Args>
auto create(size_t n, Args ... args) {
	return vector<decltype(create<T>(args...))>(n, create<T>(args...));
}
#endif
void run() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
#ifdef EZZAT
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#else
#endif
}

const ll MAX = 1e18;

int main() {
	run();
	int n;
	cin >> n;
	vector<int> v(n + 1);
	for (auto &it : v)
		cin >> it;
	vector<ll> can(n + 1);
	ll number_of_nodes = 1;
	for (int i = 0; i <= n; i++) {
		if (v[i] > number_of_nodes)
			return cout << -1, 0;
		can[i] = number_of_nodes;
		number_of_nodes = min(number_of_nodes - v[i] << 1, MAX);
	}
	ll sum = number_of_nodes = v[n];
	for (int i = n - 1; i >= 0; i--) {
		number_of_nodes = min(v[i] + number_of_nodes, can[i]);
		sum += number_of_nodes;
	}
	cout << sum << endl;
}