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>
using namespace std;
#define ll long long
#define ull unsigned ll
#define PI acos(-1)
#define vi vector<int>
#define pi pair<int, int>
#define pl pair<ll, ll>
template<typename T> using v2 = vector<vector<T>>;
template<typename T> using v3 = vector<vector<vector<T>>>;
template<class T> istream &operator>>(istream&is,vector<T>&v){for(auto &elemnt:v)is>>elemnt;return is;}
template<class T,class U> istream &operator>>(istream&is,pair<T,U>&p){is>>p.first>>p.second;return is;}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int MOD = 998244353;
const int INF = 1e9 + 200;
const ll INFLL = 2e18 + 200;

int main() {
    ll N;
    cin >> N;
    vector<ll> A(N + 1);
    cin >> A;
    ll now = 1;
    vector<ll> max_v(N + 1);
    max_v[0] = 1;
    if (N == 0 && A[0] == 1) {
        cout << 1 << endl;
        return 0;
    }
    if (A[0] != 0) {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 1; i <= N; i++) {
        now *= 2;
        if (now < 0) {
            now = INFLL;
        }
        now -= A[i];
        max_v[i] = now;
        if (now <= 0 & i != N) {
            cout << -1 << endl;
            return 0;
        }
    }
    ll ans = 0;
    now = 0;
    for (int i = N; i >= 1; i--) {
        now += A[i];
        ans += now;
        if (now > max_v[i - 1] * 2) {
            cout << -1 << endl;
            return 0;
        }
        if (now > max_v[i - 1]) {
            now = max_v[i - 1];
        }
    }
    cout << ans + 1 << endl;
}