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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

#define rep(i,n) for(int64_t i = 0;i < n;i++)

template<class T> bool chmin(T& a,T b){
    if(a > b){ a = b; return true; }else return false;
}
template<class T> bool chmax(T& a,T b){
    if(a < b){ a = b; return true; }else return false;
}

class CFolia {
public:
    void solve(std::istream& cin, std::ostream& cout) {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);

        typedef long long ll;

        int64_t n;cin >> n;
        vector<int64_t> a(n + 1);rep(i,n + 1)cin >> a[i];

        if(n == 0){
            if(a[0] != 1){
                cout << -1 << '\n';
            }else{
                cout << 1 << '\n';
            }
            return;
        }
        if(a[0] == 1){
            // THINK AFTER!!
            if(n >= 1)cout << -1 << '\n';
            else cout << 1 << '\n';
            return;
        }
        int64_t ans = 0;
        vector<int64_t> white(n + 1, 1);
        for(int i = 0;i < n;i++){
            if(2 * white[i] < a[i+1]){
                cout << -1 << '\n';
                return;
            }
            white[i+1] = 2 * white[i] - a[i+1];
        }
        white[n] = 0;
        for(int i = n - 1;i >= 0;i--){
            int64_t tmp = white[i+1] + a[i+1];
            if(white[i] > tmp) white[i] = tmp;
        }
        white[0] = 1;
        for(int i = 0;i <= n;i++) ans += white[i] + a[i];
        cout << ans<< '\n';
    }
    void main2(istream& cin,ostream& cout){

    }
};


int main() {
	CFolia solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}