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 <utility>
#include <queue>
#include <string>
#include <set>
#include <bitset>
#include <unordered_map>
#include <cmath>
#include <functional>
#include <map>
#include <cstdlib>
#include <stack>

using namespace std;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
template <class T>
using Vvec = vector<vector<T> >;
using Graph = Vvec<int>;

template <class T>
void debug(T& v)
{
    for(auto& x: v) cout << x << " ";
    cout << endl;
}

int main()
{
    int N;
    cin >> N;
    vector<ll> A(N+1), B(N+1), C(N+1);
    for(auto& x: A) cin >> x;
    if(A[0]!=0){
        cout << -1 << endl; return 0;
    }

    C[0] = 1;
    const ll infty = 1e12;
    for(int i=0; i<N; i++){
        C[i+1] = min(2*C[i] - A[i+1], infty);
        if(i!=N-1 && C[i+1] <= 0){
            cout << -1 << endl;
            return 0;
        }
    }

    B[N] = A[N];
    for(int i=N; i>0; i--){
        if((B[i]+1)/2 > C[i-1]){
            cout << -1 << endl; return 0;
        }
        B[i-1] = min(B[i], C[i-1]) + A[i-1];
    }

    ll ans = 0;
    for(int i=0; i<=N; i++) ans += B[i];
    cout << ans << endl;
}