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>
#define rep(i, n) for (int i = 0; i < (n); ++i) 
#define rrep(i, a, b) for (int i = (a); i < (b); ++i)
#define PI acos(-1)
#define pcnt __builtin_popcountll
#define rng(a) a.begin(), a.end()
#define rrng(a) a.rbegin(), a.rend()
#define sz(x) (int)(x).size()
#define v(T) vector<T>
#define vv(T) v(v(T))
#define fi first
#define se second

using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using LP = pair<ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using tl = tuple<ll, ll, ll>;

template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) {if (a > b) { a = b; return true; } return false; }

ll INF = 1001001001;
ll LINF = 1001001001001001001ll;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n; cin>>n;
    vl as(n + 1); cin>>as;
    vl cs(n + 1);

    if (as[0] > 0) {
        if (n == 0 && as[0] == 1) {
            cout<<1<<endl;
            return 0;
        }
        else {
            cout<<-1<<endl;
            return 0;
        }
    }

    cs[0] = 1;
    ll cnt = 1;
    rep(i, n) {
        if (as[i + 1] > cs[i] * 2) {
            cout<<-1<<endl;
            return 0;
        }
        cs[i + 1] = cs[i] * 2 - as[i + 1];
        cnt += cs[i] * 2;
    }
    cnt -= cs.back();
    ll temp = cs.back() + as.back();
    ll node = 0;
    ll d = n;
    while (d) {
        node += as[d];
        temp = (temp + 1) / 2;
        cnt -= max(0ll, temp - node);
        d--;
    }
    cout<<cnt<<endl;
}