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 (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;

int n;
ll ans = -1;

void dfs(int i, int leaf, vector<int> &node, vector<int> &a) {
    if(i == n-1) {
        if(node.at(i) != a.at(i)) return;
        ll tot = 0;
        rep(i, n) {
            tot += node.at(i);
        }
        ans = max(ans, tot);
        return;
    }

    int root = node.at(i) - a.at(i);
    leaf -= a.at(i);
    if(root <= 0) return;
    node.at(i+1) = min(root * 2, leaf);
    dfs(i+1, leaf, node, a);
}

int main() {
    cin >> n;
    n++;
    vector<int> a(n);
    int leaf = 0;
    rep(i, n) {
        cin >> a.at(i);
        leaf += a.at(i);
    }

    vector<int> node(n);
    node.at(0) = 1;
    dfs(0, leaf, node, a);
    cout << ans << endl;
    return 0;
}