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;
using ll = long long;
using uint = unsigned int;
#define rep(i,n) for(int i=0;i<int(n);i++)
#define rep1(i,n) for(int i=1;i<=int(n);i++)
#define per(i,n) for(int i=int(n)-1;i>=0;i--)
#define per1(i,n) for(int i=int(n);i>0;i--)
#define all(c) c.begin(),c.end()
#define si(x) int(x.size())
#define pb emplace_back
#define fs first
#define sc second
template<class T> using V = vector<T>;
template<class T> using VV = vector<vector<T>>;
template<class T,class U> void chmax(T& x, U y){if(x<y) x=y;}
template<class T,class U> void chmin(T& x, U y){if(y<x) x=y;}
template<class T> void mkuni(V<T>& v){sort(all(v));v.erase(unique(all(v)),v.end());}
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
	return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
	o<<"{";
	for(const T& v:vc) o<<v<<",";
	o<<"}";
	return o;
}
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

#ifdef LOCAL
#define show(x) cerr << "LINE" << __LINE__ << " : " << #x << " = " << (x) << endl
void dmpr(ostream& os){os<<endl;}
template<class T,class... Args>
void dmpr(ostream&os,const T&t,const Args&... args){
	os<<t<<" ~ ";
	dmpr(os,args...);
}
#define shows(...) cerr << "LINE" << __LINE__ << " : ";dmpr(cerr,##__VA_ARGS__)
#define dump(x) cerr << "LINE" << __LINE__ << " : " << #x << " = {";  \
	for(auto v: x) cerr << v << ","; cerr << "}" << endl;
#else
#define show(x) void(0)
#define dump(x) void(0)
#define shows(...) void(0)
#endif

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);		//DON'T USE scanf/printf/puts !!
	cout << fixed << setprecision(20);

	int N; cin >> N;
	V<ll> A(N+1); rep(i,N+1) cin >> A[i];

	ll L = 1, R = 1;
	V<ll> Ls(N+1),Rs(N+1);
	const ll inf = 1e18;
	rep(i,N+1){
		Ls[i] = L, Rs[i] = R;
		if(R < A[i]){
			cout << -1 << endl;
			return 0;
		}
		L -= A[i], R -= A[i];
		if(L<0) L = 0;
		R *= 2;
		chmin(R,inf);
	}
	if(L == 0){
		ll ans = 0;
		ll x = 0;
		per(i,N+1){
			x = min(Rs[i],x+A[i]);
			ans += x;
		}
		cout << ans << endl;
	}else{
		cout << -1 << endl;
		return 0;
	}
}