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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
long long minim(long long a, long long b) { return a < b ? a : b; }
int main(){
    long long n,i,ans=0,leafmax[100010]={},leafmin[100010]={},nodemax[100010]={},a[100010]={};
    scanf("%lld",&n);
    scanf("%lld",&a[0]);
    if(a[0]>=1 && n>=1){
        printf("-1");
        return 0;
    }
    leafmax[0]=1;
    for(i=1; i<=n; i++){
        scanf("%lld",&a[i]);
        leafmax[i]=2*(leafmax[i-1]-a[i-1]);
        if(leafmax[i]<a[i] ){
            printf("-1");
            return 0;
        }
    }
    nodemax[n]=a[n];
    for(i=n-1; i>=0; i--){
        nodemax[i]=minim(nodemax[i+1]+a[i],leafmax[i]);
    }
    for(i=0; i<=n; i++){
        ans+=nodemax[i];
    }
    printf("%lld",ans);
    return 0;
}