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 <stdio.h>

long min(long a,long b){
  if(a>=b){return b;}
  else{return a;}
}

int main(){
  long n,ans=1;
  scanf("%ld",&n);
  long a[n+2],b[n+2],sum[n+2];
  for(long i=0;i<=n;i++)scanf("%ld",&a[i]);
  
  if(a[0]>0){printf("-1");return 0;}
  
  b[1]=2;
  sum[0]=0;
  
  for(long i=1;i<=n;i++){
    sum[0]+=a[i];
  }
  
  for(long i=1;i<=n;i++){
    if(b[n]<a[n]){printf("-1");return 0;}
    
    sum[i]=sum[i-1]-a[i];
    b[n+1]=2*(b[n]-a[n]);
  }
  
  for(long i=1;i<=n;i++){
    ans+=min(b[i]-a[i],sum[i]);
    ans+=a[i];
  }
  
  printf("%ld",ans);
}