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>
using namespace std;
int main(){
unsigned long long n,sum=0;cin>>n;n++;int a[n+5];for(unsigned long long i=1;i<=n;i++){cin>>a[i];}int b[100005]={0},c[100005]={0};b[1]=1,c[n]=a[n];
for(unsigned long long i=2;i<=n;i++){b[i]=(b[i-1]-a[i-1])*2;}
for(unsigned long long i=n-1;i>=1;i--){c[i]=c[i+1]+a[i];}
for(unsigned long long i=1;i<=n;i++){sum+=min(b[i],c[i]);}
for(unsigned long long i=1;i<=n;i++){if(a[i]>b[i]){return cout<<"-1"<<endl,0;}}
cout<<sum<<endl;return 0;}