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;
typedef long long ll;

#define INX(x) (cin >> x)
#define INXV(vec, cnt) for(ll i; i < cnt; i++) INX(vec[i])
#define OUTX(x) (cout << x << endl)
#define OUTXV(vec) for(ll i; i < (ll)vec.size(); i++) OUTX(vec[i])

#define ADD emplace_back
#define MP make_pair

int main()
{
    ll N;
    INX(N);

    vector<ll> A(N+1);
    for(ll i = 0; i < N+1; i++)
    {
        cin >> A[i];
    }
    if(A[0] > 0 || A[N] > pow(2, N))
    {
        OUTX(-1);
        return 0;
    }
    
    vector<ll> sumA(N+1);
    sumA[N] = A[N];
    for(ll i = N - 1; i > 0; i--)
    {
        sumA[i] = A[i] + sumA[i + 1];
    }

    bool isOK = true;
    vector<ll> vertex(N+1);
    vertex[0] = 1;
    ll result = 1; //各depthの頂点の合計
    for (ll depth = 1; depth <= N; depth++)
    {
        vertex[depth] = min(2 * (vertex[depth - 1] - A[depth - 1]), sumA[depth]);
        result += vertex[depth];
    }
    OUTX((isOK ? result : -1));
    

    return 0;
}