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)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define eb emplace_back

using namespace std;
long long n,ans,bef,arr[100005],bsr[100005];
bool sabi;
int main(){
	cin>>n;
	for(int i =0;i<=n;i++){
		cin>>arr[i];
	}
	sabi=1;
	if(arr[0]>0)sabi=0;
	for(int i =0;i<=n;i++){
		if(i==0)bsr[i]=1;
		else{
			bsr[i]=bsr[i-1]*2-arr[i];

		}
		if(i<n&&bsr[i]<=0)sabi=0;
		if(bsr[i]<0)sabi=0;
	}
	for(int i =n;i>=0;i--){
		if(i==n){
			bef=arr[i];
			ans+=bef;
		}
		else{
			if((bef+1)/2>bsr[i])sabi=0;
			bef= min(bef,bsr[i])+arr[i];
			ans+=bef;
		}
	}
	if(sabi)cout<<ans<<endl;
	else cout<<-1<<endl;

}