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 <vector>
#include <algorithm>
#include <set>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
    vector<ll> a(n+1);
    bool ok=1;
    ll res=0;
    for(int i=0;i<n+1;i++){
        cin >> a[i];
    }
    if(a[0]==1){
        if(n==0){
            printf("1\n");
        }
        else{
            printf("-1\n");
        }
        return 0;
    }
    vector<ll> mi(n+1),ma(n+1);
    for(int i=n;i>=0;i--){
        if(i==n){
            mi[i]=ma[i]=a[i];
        }
        else{
            ma[i]=ma[i+1]+a[i];
            mi[i]=a[i]+(mi[i+1]+1)/2;
        }
    }
    /*for(int i=0;i<=n;i++){
        cout << mi[i] << " " << ma[i] << endl;
    }*/
    res++;
    ll now=1;
    for(int i=1;i<=n;i++){
        if(mi[i]<=2*now){
            now=min(2*now,ma[i]);
            //cout << now << endl;
            res+=now;
            now-=a[i];
        }
        else{
            ok=0;
            break;
        }
    }
    if(ok){
        printf("%lld\n",res);
    }
    else{
        printf("-1\n");
    }
}