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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include <bits/stdc++.h>

using namespace std;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n;
	cin >> n;
	vector<long long> a(n+1);
	for(int i=0;i<=n;i++) cin >> a[i];
	
	long long noko=1;
	for(int i=0;i<=n;i++){
		if(noko<a[i]){
			cout << -1 << "\n";
			return 0;
		}
		noko-=a[i];
		if(noko<(1LL<<55)) noko*=2;
	}
	
	long long res=0;
	vector<long long> son(n+1);
	noko=1;
	for(int i=0;i<=n;i++){
		son[i]=noko;
		noko-=a[i];
		if(noko<(1LL<<55)) noko*=2;
	}
	
	long long v=0;
	for(int i=n;i>=0;i--){
		v+=a[i];
		v=min(v,son[i]);
		res+=v;
	}
	cout << res << "\n";
	return 0;
}