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 i64 = long long;
using P = pair<i64, i64>;

#define overload3(_1, _2, _3, name, ...) name
#define rep1(i, n) for(i64 i = 0LL; i < (n); ++i)
#define rep2(i, a, b) for(i64 i = (a); i < (b); ++i)
#define rep(...) overload3(__VA_ARGS__, rep2, rep1)(__VA_ARGS__)
#define all(v) v.begin(), v.end()


void solve(long long N, std::vector<long long> A){
    vector<i64> table(N + 1);
    for(int i = N - 1; i >= 0; --i){
        table[i] = table[i + 1] + A[i + 1];
    }
    i64 ans = 0, cur = 1;
    if(N == 0){
        if(A[0] == 1) cout << 1 << endl;
        else cout << -1 << endl;
        return;
    }
    rep(i, N + 1){
        ans += cur;
        cur = min((cur - A[i]) * 2, table[i]);
        if(i != N && cur <= 0){
            cout << -1 << endl;
            return;
        }
    }
    cout << ans << endl;
}

struct IoSetup {
    IoSetup() {
        // cin.tie(nullptr);
        // ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;

int main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> A(N-0+1);
    for(int i = 0 ; i < N-0+1 ; i++){
        scanf("%lld",&A[i]);
    }
    solve(N, std::move(A));
    return 0;
}