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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<ll>;
using vvi = vector<vi>;



int main() {
    int n;
    cin >> n;
    vi a(n+1);
    rep(i, n+1) {
        cin >> a[i];
    }
    if(!n) {
        if(a[0]) {
            cout << 1 << endl;
        } else {
            cout << 0 << endl;
        }
        return 0;
    }

    vi sum(n+1);
    sum[n] = a[n];
    rep(i, n){
        sum[n-i-1] = sum[n-i]+a[n-i-1];
    }
    bool x = true;
    vi vec(n+1);
    vec[0] = 1;
    if(a[0] == 1 && n) x = false;
    rep(i, n) {
        vec[i+1] = vec[i]-a[i];
        if(vec[i+1]*2-a[i+1] <= 0) {
            if(i != n-1) x = false;
            if(i == n-1 && vec[i+1]*2-a[i+1] < 0) x = false;
        }
        vec[i+1] *= 2;
        if(vec[i+1] > sum[i+1]) {
            vec[i+1] = sum[i+1];
        }
    }

    ll ans = 0;
    rep(i, n+1) {
        //cout << sum[i] << endl;
        ans += vec[i];
    }
    if(!x) {
        cout << -1 << endl;
        return 0;
    }
    cout <<ans << endl;

    
}