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;
const ll INF = 1e18;
const int inf = 1e9;
double pi = 3.14159265359;
#define rep(i, a, b) for (int i = a; i < b; i++)
#define per(i, b, a) for (int i = a - 1; i >= b; i--)
using Graph = vector<vector<int>>;
using pint = pair<int, int>;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dxx[8] = {1, 1, 1, 0, 0, -1, -1, -1}, dyy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};

int main() {
    int n;
    cin >> n;
    int a[n + 1];
    rep (i, 0, n + 1) cin >> a[i];
    if (a[0] > 0) {
        cout << -1 << "\n";
        return 0;
    }
    ll ans = 0;
    ll s[n + 1]; //要素の最大の数
    rep (i, 0, n + 1) {
        if (i == 0) s[i] = 1;
        else {
            s[i] = (s[i - 1] - a[i - 1]) * 2;
        }
        if (s[i] < 0) {
            cout << -1 << "\n";
            return 0;
        }
    }
    ll b[n + 1];
    rep (i, 0, n + 1) b[i] = 0;
    per (i, 0, n + 1) {
        if (i == n) b[i] = a[i];
        else {
            b[i] = b[i + 1] + a[i];
            if (b[i] > s[i]) b[i] = s[i];
        }
        ans += b[i];
        // cout << b[i] << "\n";
    }
    cout << ans << "\n";
}