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)

/* atcoder */
#include "bits/stdc++.h"
using namespace std;

/* typedef */
/* long系 */
typedef long long ll;   // long long
typedef long double ld; // long double
/* vector系 */
typedef vector<int> vi;                    // intのvector
typedef vector<double> vd;                 // doubleのvector
typedef vector<ll> vll;                    // llのvector
typedef vector<ld> vld;                    // ldのvector
typedef vector<pair<int, int>> vi_i;       // i_iのvector
typedef vector<pair<double, double>> vd_d; // d_dのvector
typedef vector<pair<ll, ll>> vl_l;         // l_lのvector
typedef vector<pair<ld, ld>> vld_ld;       // ld_ldのvector
typedef vector<string> vs;                 // stringのvector
/* pair系 */
typedef pair<int, int> i_i;       // int同士のpair
typedef pair<double, double> d_d; // double同士のpair
typedef pair<ll, ll> l_l;         // ll同士のpair
typedef pair<ld, ld> ld_ld;       // ld同士のpair

/* define */
#define all(v) v.begin(), v.end() // コンテナ全体
#define PI (acos(-1))             // 円周率pi

/* stringstream */
stringstream ss;
stringstream init_ss; // save default formatting

int nsl[100000 + 1];
int aa[100000 + 1];

bool check(ll i)
{
    ll d = nsl[i - 1] - (aa[i] + nsl[i]);
    if (d > 0)
    {
        nsl[i - 1] -= d;
    }
    else
    {
        return true;
    }
    if (i - 1 == 0)
    {
        return true;
    }
    else
    {
        return check(i - 1);
    }
}

/* main */
void Main(void)
{
    ll n;
    cin >> n;
    n++;
    vi a;
    a.resize(n);
    for (ll rv = 0; rv < n; rv++)
    {
        cin >> a[rv];
    }
    if (a[0] == 2 || (a[0] == 1 && n >= 1))
    {
        cout << -1 << endl;
        return;
    }
    else if (a[0] == 1 && n == 0)
    {
        cout << 1 << endl;
        return;
    }
    for (ll i = 0; i < n; i++)
    {
        aa[i] = a[i];
    }
    nsl[0] = 1;
    ll ans = 0, ns = 1;
    for (ll i = 1; i < n; i++)
    {
        if (ns * 2 < a[i])
        {
            cout << -1 << endl;
            return;
        }
        ns = ns * 2 - a[i];
        nsl[i] = ns;
        if (i == n - 1)
        {
            nsl[i] = 0;
            bool f = check(i);
            if (f == false)
            {
                cout << -1 << endl;
                return;
            }
        }
    }
    for (ll i = 0; i < n; i++)
    {
        ans = ans + nsl[i] + aa[i];
    }
    cout << ans << endl;
}

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

    return 0;
}