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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i,x,y) for(ll i=(x);i<(y);i++)
#define rrep(i,x,y) for(ll i=(ll)(y)-1;i>=(x);i--)
#define all(x) (x).begin(),(x).end()
#ifdef LOCAL
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
#define itrdebug(x) cerr << #x << " "; for (auto & el : (x)) {cerr << (el) << " ";} cerr << endl
#define dassert(...) assert(__VA_ARGS__)
#else
#define debug(x)
#define itrdebug(x)
#define dassert(...)
#endif
#define int long long

typedef long long ll;
const ll MOD = 1e9 + 7;
const long double EPS = 1e-8;

class BIT
{
public:
    vector<int> bit;
    int M;

    BIT(int M):
            bit(vector<int>(M+1, 0)), M(M) {}

    int sum(int i) {
        if (!i) return 0;
        return bit[i] + sum(i-(i&-i));
    }

    void add(int i, int x) {
        if (i > M) return;
        bit[i] += x;
        add(i+(i&-i), x);
    }
};

void solve(long long N, std::vector<long long> A){
    if (N == 0) {
        cout << (A[0] == 1 ? 1 : -1) << endl;
        return;
    }
    if (A[0] != 0) {
        cout << -1 << endl;
        return;
    }
    BIT bit(N+2);
    bit.add(1,2);
    bit.add(2,-2);
    ll answer = 1;
    int top = 1;
    rep(i,1,N+1) {
        if (A[i] == 0) continue;
        while (A[i] > 0) {
            while (top <= i && bit.sum(top) <= 0) {
                top++;
            }
            if (top > i) {
                cout << -1 << endl;
                return;
            }
            ll cnt = min<ll>(bit.sum(top), A[i]);
            answer += cnt * (i - top + 1);
            bit.add(top, -cnt);
            bit.add(top+1, 2*cnt);
            bit.add(i+1,-cnt);
            A[i] -= cnt;
        }
    }
    cout << answer << endl;
}

signed main(){
    // ios_base::sync_with_stdio(false);
    // cin.tie(NULL);
    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;
}