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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>  
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
ll MOD = 1e9+7;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 
ll mpow(ll a, ll b){
	if(a==0) return 0;
	if(b >= (MOD-1)) b %= (MOD-1);
	if(b==0) return 1;
	ll t1 = mpow(a,b/2);
	t1 *= t1;
	t1 %= MOD;
	if(b%2) t1 *= a;
	t1 %= MOD;
	return t1;
}
 
ll mpow(ll a, ll b, ll p){
	if(a==0) return 0;
	if(b==0) return 1;
	ll t1 = mpow(a,b/2,p);
	t1 *= t1;
	t1 %= p;
	if(b%2) t1 *= a;
	t1 %= p;
	return t1;
}
 
ll modinverse(ll a, ll m){
	ll m0 = m; 
    ll y = 0, x = 1; 
    if (m == 1) return 0; 
    while (a > 1){
        ll q = a / m; 
        ll t = m; 
        m = a % m, a = t; 
        t = y; 
        y = x - q * y; 
        x = t; 
    }
    if (x < 0) x += m0; 
    return x; 
}

mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
 
ll range(ll l, ll r){
	return l + mt()%(r-l+1);
}
 
ll rev(ll v){
	return mpow(v,MOD-2);
}

void solve(){
	ll n;
	cin >> n;
	n++;
	ll a[n];
	forn(i,n) cin >> a[i];
	ll b[n];
	ll pr = 1;
	forn(i,n){
		b[i]=min(pr-a[i],(ll)INF);
		if(b[i]<0){
			cout << -1 << ln;
			return;
		}
		pr = 2*b[i];
	}
	b[n-1]=0;
	rforn(i,n-2){
		b[i]=min(b[i],b[i+1]+a[i+1]);
	}
	ll ans = 0;
	forn(i,n) ans += b[i]+a[i];
	cout << ans << ln;
}
 
int main()
{
	fast_cin();
    ll t=1;
    // cin >> t;
    forn(i,t) {
    	// cout << "Case #" << i+1 << ": ";
    	solve();
    }
    return 0;
}