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>
//#include <boost/multiprecision/cpp_int.hpp>
//namespace mp = boost::multiprecision;

using namespace std;

const double PI = 3.14159265358979323846;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
typedef pair<ll, ll> P;
const ll INF = 10e17;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

double equal(double a, double b) {
    return fabs(a - b) < DBL_EPSILON;
}

std::istream &operator>>(std::istream &in, set<int> &o) {
    int a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
//const ll mod = 1e10;

typedef priority_queue<ll, vector<ll>, greater<ll> > PQ_ASK;

int main() {
    int n;
    cin >> n;
    n++;
    vector<ll> v(n);
    rep(i, n) cin >> v[i];

    vector<ll> u_min(n, 0), u_max(n, 0);

    for (int i = n - 2; i >= 0; i--) {
        u_min[i] = (u_min[i + 1] + v[i + 1] + 1) / 2;
        u_max[i] = u_max[i + 1] + v[i + 1];
    }


    vector<ll> d_min(n, 0), d_max(n, 0);
    d_min[0] = 1 - v[0];
    d_max[0] = 1 - v[0];
    for (int i = 1; i < n; i++) {
        ll mi = d_min[i - 1] - v[i];
        ll ma = d_max[i - 1] * 2 - v[i];
        d_min[i] = max(mi, u_min[i]);
        d_max[i] = min(ma, u_max[i]);

        if (d_max[i] - d_min[i] < 0) {
            cout << -1 << endl;
            ret();
        }
    }

    ll ans = accumulate(d_max.begin(), d_max.end(), 0ll) + accumulate(v.begin(), v.end(), 0ll);



    cout << ans << endl;


}