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 <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;
using LLONG = long long;
const LLONG MOD = 1000000007;

int main()
{
    int N; cin >> N;
    if (N == 0)
    {
        // 根ノードのみの場合
        int a0; cin >> a0;
        cout << (a0 == 1 ? 1 : -1) << endl;
        return 0;
    }
    vector<LLONG> as(N + 1); for (auto &ai : as) cin >> ai;
    if (as[0] != 0)
    {
        // 深さN>0のとき，A0はゼロでなければならない
        cout << -1 << endl;
        return 0;
    }
	vector<LLONG> sums(N + 1);	// sums[i] = Ai + ... + AN
	LLONG sum = 0;
	for (int i = N; i >= 0; --i)
	{
		sums[i] = sum;
		sum += as[i];
	}
	// 各深さdにおいて葉でない頂点数bs[d]を計算する
	vector<LLONG> bs(N + 1);
	bs[0] = 1;
	for (int d = 1; d <= N; ++d)
	{
		bs[d] = min(sums[d], bs[d - 1] * 2 - as[d]);
	}
	LLONG ans = 0;
	for (int d = 0; d <= N; ++d)
	{
		ans += as[d] + bs[d];
	}
	cout << ans << endl;
}