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>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>

int main()
{
  int n, i;
  scanf("%d", &n);
  int a[n];
  int sum = 0;
  for(i = 0; i <= n; i++){
    scanf("%d", &a[i]);
    sum = sum + a[i];
  }
  int eda = 1, result = 1;
  int frag = 0;
  if(a[0] != 0){
    printf("-1\n");
    return 0;
  }else{
    for(i = 1; i <= n; i++){
      if(eda*2 <= sum){
        if(eda*2 <= a[i]){
          frag = 1;
          break;
        }
        eda = eda*2 - a[i];
      }else{
        eda = sum - a[i];
      }
      result = result + eda + a[i];
      sum = sum - a[i];
    }

    if(frag == 0){
      printf("%d\n", result);
    }else{
      printf("-1\n");
    }
    return 0;
  }
}