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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#define N 300000
int main()
{

    int n;
    long long lef[N];
    int flag = 0;
    long long node[N];
    long long hutosa[N];
    long long x;

    scanf("%d", &n);

    for (int i = 0; i <=n; i++)
    { //道を読み取る
        scanf("%lld", &lef[i]);
    }
    //ここまでは良い

  
    node[0] = 1;

    if (n == 0)
    {//根がたくさん
        if(lef[0]==1){
            hutosa[0] = 1;
        }
        else flag =1;
    }
    else {   
        for (int i = 0; i <= n; i++)
        {
            node[i + 1] = (node[i] - lef[i])*2;

           // printf("%ld\n", node[i]);

            if (i != n && node[i + 1] <= 0)
            {
                flag = 1; //damejyatta
            }
          //  printf("%d ノードにて太さ%ld\n ",i,hutosa[i]);
        }
    }
    if (flag != 1){
    hutosa[n] = lef[n];
    hutosa[0] = 1;

    hutosa[n] = lef[n];
    for (int i = n - 1; i >= 0; i--)
    { //道を読み取る
        x = hutosa[i + 1];

        if (node[i] < hutosa[i + 1])
        {
            x = node[i];
        }

        if (node[i] < lef[i])
        {
            lef[i] = node[i];
        }


        if (node[i] < lef[i] + x)
        {
            hutosa[i] = node[i];
        }
        else
            hutosa[i] = lef[i] + x;
    }


    node[n] = 0;

    }
    if(flag==1)
        printf("-1");
    else {
        long long ans = 0;
        for (int i = 0; i <= n; i++){
            ans += hutosa[i];
        }
        printf("%lld",ans);
    }

    printf("\n");
}