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 <math.h>

#define N 200000

int main (void)
{
    long long a[N], node2[N] = {0}, node0[N] = {0};
    long long n, i, ans = 0;

    scanf("%lld", &n);

    for (i = 0; i < n + 1; i++) {
        scanf("%lld", &(a[i]));
    }

    if (n == 0 && a[0] == 1) {
        puts("1");
    } else if (n == 0 && a[0] != 1) {
        puts("-1");
    }

    if (a[0] != 0 || a[n] == 0) {
        puts("-1");
        return 0;
    }

    node2[0] = 1;
    for (i = 1; i < n + 1; i++) {
        node2[i] = node2[i - 1] * 2 - a[i];
        if (node2[i] > 100000000) node2[i] = 100000000;

        if (node2[i] < 0) {
            puts("-1");
            return 0;
        }

        node0[i] = a[i];
    }

    for (i = n; i > 0; i--) {
        ans += node0[i];
        if (node0[i] < node2[i - 1]) {
            node0[i - 1] += node0[i];
        } else {
            node0[i - 1] += node2[i - 1];
        }
    }

    ans += 1;

    printf("%lld", ans);

    return 0;
}