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 <cstdint>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define rep(i, n) for(int(i) = 0; (i) < (n); (i)++)
#define FOR(i, m, n) for(int(i) = (m); (i) < (n); (i)++)
#define All(v) (v).begin(), (v).end()
#define pb push_back
#define MP(a, b) make_pair((a), (b))
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const ll INF = 1 << 30;
const ll LINF = 1LL << 60;
const int MOD = 1e9 + 7;

int main() {
    int N;
    cin >> N;
    vector<int> A(N + 1);
    rep(i, N + 1) cin >> A[i];
    ll num = 1;
    ll res = 1;
    vector<ll> lim(N + 1);
    rep(i, N + 1) {
        lim[i] = num;
        num -= A[i];
        if(i != N && num <= 0) {
            cout << -1 << endl;
            return 0;
        }
        num = min(num * 2, INF);
    }

    ll cnt = 0;
    ll ans = 0;
    for(int i = N; i >= 0; i--) {
        cnt += A[i];
        if(cnt > lim[i])
            cnt = lim[i];
        ans += cnt;
    }

    cout << ans << endl;
    return 0;
}