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;

using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using P = pair<ll, ll>;
template <typename T> using pq = priority_queue<T>;
template <typename T> using minpq = priority_queue<T, vector<T>, greater<T>>;
template <typename T, typename K> using ump = unordered_map<T, K>;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll mod = 1000000007;
const ll inf = ll(1e9);
const ll e5 = ll(1e5);
const ll ll_inf = ll(1e9) * ll(1e9);

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for (ll i = ll(n - 1); i >= 0; i--)
#define each(i, mp) for (auto &i : mp)
#define eb emplace_back
#define F first
#define S second
#define all(obj) (obj).begin(), (obj).end()

template <class T> ostream &operator<<(ostream &out, const vector<T> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << ' ';
    return out;
}

template <class T> istream &operator>>(istream &in, vector<T> &list) {
    ll n = list.size();
    rep(i, n) in >> list[i];
    return in;
}

template <class T> ostream &operator<<(ostream &out, const vector<vector<T>> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << '\n';
    return out;
}

/* ------------- ANSWER ------------- */
/* ---------------------------------- */

void solve() {
    ll n;
    cin >> n;
    vl a(n + 1);
    cin >> a;
    vl ne_max(n + 1);
    rep(i, n + 1) {
        ll maxv = 0;
        if (i > 50) maxv = ll_inf;
        else maxv = pow(2, i);
        if (i >= 1) {
            ne_max[i] = max(0LL, min(ne_max[i - 1] * 2 - a[i], maxv - a[i]));
        } else {
            ne_max[i] = max(0LL, maxv - a[i]);
        }
    }
    if (n == 0 && a[0]== 0 && a[1]==0) {
        cout << 1 << '\n';
        return;
    }
    if (a.back() == 0) {
        cout << -1 << '\n';
        return;
    }
    ll now_v = a.back();
    ll ans = now_v;
    for (ll i = n - 1; i >= 0; --i) {
        if (now_v > ne_max[i] * 2) {
            cout << -1 << '\n';
            return;
        }
        now_v = min(ne_max[i] + a[i], now_v + a[i]);
        ans += now_v;
    }
    if (ans <= 0) {
        cout << -1 << '\n';
    } else cout << ans << '\n';
}

int main() {
#ifdef MY_DEBUG
    while (true) {
#endif
        solve();
#ifdef MY_DEBUG
    }
#endif
    return 0;
}