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<bits/stdc++.h>
 
#define rep(i,n) for(ll i = 0;i < n;++i)
#define all(v) v.begin(),v.end()
using namespace std;
using ll = long long;
typedef pair<int,int> P;
 
const int INF = 1001001001;
const long double PI = (acos(-1));
const int mod = 1e9+7;
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;
    vector<int> a(n+1);
    rep(i,n+1){
        cin >> a[i];
    }
    ll sum = 0;
    auto tmp = a;
    for(int i = n;i > 0;--i){
        tmp[i-1] += (tmp[i]+1)/2;
    }
    if(tmp[0] >= 2) {
        cout << -1 << endl;
        return 0;
    }
    vector<ll> max(n+1,1);
    for(int i = 1;i <= n;++i){
        max[i] = (2*max[i-1])-a[i];
    }
    ll ans = 0;
    for(ll i = n;i > 0;--i){
        if(a[i] <= max[i-1]){
            ans += a[i];
            a[i-1] += a[i];
        }else {
            ans += a[i];
            a[i-1] += (a[i]+1)/2;
        }
    }
    ans++;
    cout << ans << endl;
    cout << endl;
    return 0;
}