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<iostream>
#include<string>
#include<vector>
#include<algorithm>//sort,reverse
#include<utility>//pair
#include<queue>//queue,priority_queue
#include<deque>//double_ended queue
#include<map>//map
using namespace std;
typedef long long LL; 
typedef pair<LL,LL> P;//Pでpair<-,->を表す。
const LL MAX=2000010;
const LL MOD=1000000007;
typedef map<LL,LL>::iterator Itr;

int main()
{ 
    LL n;
    cin>>n;
    LL a[n+1];
    for(int i=0;i<=n;i++){
        cin>>a[i];
    }
    LL b[n+1];
    b[0]=1-a[0];
    if(b[0]<0){cout<<-1<<endl;return 0;}
    for(int i=1;i<=n;i++){
        b[i]=b[i-1]*2-a[i];
        if(b[i]>1000000000000){b[i]=1000000000000;}
        if(b[i]<0){cout<<-1<<endl;return 0;}
    }
    LL ans=0;
    LL cnt=0;
    for(int i=n;i>=0;i--){
        if(cnt>b[i]){cnt=b[i]+a[i];}
        else{cnt=cnt+a[i];}
        ans=ans+cnt;
    }
    cout<<ans<<endl;
    return 0;
}