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>
using namespace std;

int main(){
	cin.tie(0),ios::sync_with_stdio(false);
	int n; cin>>n;
	vector<int64_t>a(n+1); for(auto&&i:a)cin>>i;
	vector<pair<int64_t,int64_t>>b(n+1);
	for(int i=0;i<n;++i)b.at(i).second=a.at(i);
	b.back()={a.back(),a.back()};
	bool yes=true;
	for(auto i=b.rbegin()+1;i!=b.rend();++i){
		i->first=((i-1)->first+1)/2;
		i->second+=(i-1)->second;
		if(i->first>i->second)yes=false;
	}
	int64_t ans=0,j=1;
	for(int i=0;i<=n;++i){
		if(j<b.at(i).first)yes=false;
		ans+=min(j,b.at(i).second);
		if(j<=4e18){
			j-=a.at(i);
			j*=2;
		}
	}
	cout<<(yes?ans:-1)<<"\n"s;
}