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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }

const int maxn = 1e6 + 11;

vector<ll> v(maxn);
vector<ll> mostposs(maxn);

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    int n;
    cin >> n;

    for (int i = 0; i <= n; i++) {
        cin >> v[i];
    }

    ll activep = 1LL;
    for (int i = 0; i <= n; i++) {
        if (v[i] > activep) out(-1);
        activep -= v[i];
        activep *= 2LL;
    }

    for (int i = n; i >= 0; i--) {
        mostposs[i] = v[i] + mostposs[i+1];
    }

    activep = 1LL;
    ll ans = 1LL;
    for (int i = 1; i <= n; i++) {
        activep *= 2LL;
        // if (v[i] > activep) out(-1);
        ans += v[i]; // add leaves
        activep -= v[i]; // minus leaves bc they can't be parents to next level
        activep = min(mostposs[i+1],activep); // bound # of node that will be parents to next level
        ans += activep; // add parents of this level
    }
    
    out(ans);
    
    return 0;
}