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 a[101010];
long long int num[101010];

int min(long long int a, long long int b){
    if(a < b) return a;
    else return b;
}

int main(){
    int n; scanf("%d",&n);
    for(int i=0; i<=n; i++) scanf("%d",&a[i]);
    num[0] = 1;
    for(int i=0; i<=n; i++){
        if(num[i] < a[i]){
            printf("-1\n");
            return 0;
        }
        if(i == n) break;
        num[i] -= a[i];
        num[i+1] = min(1LL<<60, num[i] * 2);
    }
    long long int ans = a[n];
    long long int now = a[n];
    for(int i=n-1; i>=0; i--){
        ans += min(num[i], now) + a[i];
        now += a[i];
    }
    printf("%lld\n",ans);
    return 0;
}