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>

using namespace std;
typedef long long ll;

#define REP(i,n) for(int i=0; i<int(n); i++)
#define FOR(i,m,n) for(int i=int(m); i<int(n); i++)
#define ALL(obj) (obj).begin(),(obj).end()
#define VI vector<int>
#define VP vector<pair<int,int>>
#define VPP vector<pair<int,pair<int,int>>>
#define VLL vector<long long>
#define VVI vector<vector<int>>
#define VVLL vector<vector<long long>>
#define VC vector<char>
#define VS vector<string>
#define VVC vector<vector<char>>
#define VB vector<bool>
#define VVB vector<vector<bool>>
#define fore(i,a) for(auto &i:a)
typedef pair <int, int> P;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const int  INF = 1 << 30;
const ll INFL = 1LL << 60;
const ll mod = 998244353;




int main() {

	int n;
	cin >> n;
	VLL a(n + 1);
	REP(i, n + 1)cin >> a[i];
	VLL b(n + 1, INFL);
	if (n != 0 && a[0] != 0) {
		cout << -1 << endl;
		return 0;
	}
	if (n == 0) {
		if (a[0] == 1)cout << 1 << endl;
		else cout << -1 << endl;
		return 0;
	}
	b[0] = 1;
	FOR(i, 1, n + 1) {
		b[i] = b[i - 1] * 2 - a[i];
		if (b[i] < 0) {
			cout << -1 << endl;
			return 0;
		}
	}



	reverse(ALL(a));
	reverse(ALL(b));

	REP(i, n + 1)b[i] += a[i];




	ll ans = 1;
	ll cur = 0;

	REP(i, n) {
		cur += a[i];
		cur = min(cur, b[i]);
		ans += cur;
	}
	cout << ans << endl;




}