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>
int main(){
    int a,b[100000]={},eda=2,ans=1;
  scanf("%d",&a);
  for(int i=0;i<=a;i++){
    scanf("%d",&b[i]);
  }
    
  if(a==0&&b[0]!=1) {
   printf("-1");
    return 0;
  }
  
  if(a!=0&&b[0]!=0) {
   printf("-1");
    return 0;
  }
  
  if(a==2&&b[1]==0){
    if (b[2]==0||b[2]==1){
    printf("%d",2+b[2]);
    return 0;
    }
    else{
   printf("-1");
    return 0;
    }
  }
  
  for(int i=1;i<a;i++){
    eda*=2;
    if(0<=b[i]<=eda/2){
      eda/=2;
     ans-=eda;
      eda/=2;
      ans+=eda+b[i];
      eda*=2;
    }
   else{eda-=2*b[i];
    if(eda<0) {
   printf("-1");
    return 0;
    }
    if(eda==0)
      eda++;
      ans+=eda;
       }
  }
    
  if(eda/2<=b[a]&&b[a]<=eda){
  ans+=b[a];
  }
  else{
   printf("-1");
    return 0;
  }
  printf("%d",ans);
  return 0;
}