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 ld = long double;

#define DUMP(x) cout << #x << " = " << (x) << endl;
#define REP(i, n) for(size_t i = 0; i < (n); i++)
#define REPR(i, n) for(size_t i = (n); i >= 0; i--)
#define FOR(i, a, b) for(size_t i = (a); i < (b); i++)
#define FOREACH(x, a) for(auto& (x) : (a))

const int INF = 1e9;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);

    int N; cin >> N;
    int A[N + 1];           // 各depthにおける葉の数を表す
    int B[N + 1];           // 各depthにおいて、葉以外の頂点の最大数を表す 
    bool is_exist = true;
    REP(i, N + 1) {
        cin >> A[i];
        if (i == 0) B[i] = 1 - A[i];
        else {
            B[i] = B[i - 1] * 2 - A[i];
        }
        if (B[i] < 0) is_exist = false;
    }

    int ans; 
    if (is_exist) {
        ans = 0;
        int v = 0;
        for (int i = N; i >= 0; i--) {
            if ((v + 1) / 2 > B[i]) {
                ans = -1;
                break;
            }
            if (v > B[i]) v = B[i];
            v += A[i];
            ans += v;
        }
    } else {
        ans = -1;
    }
    cout << ans << "\n";
}