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;
typedef long long ll;
typedef unsigned long long ull;
ull n,a[100010],ans;

ull dfs(ull deep,ull ent){
	if(ent<a[deep]||ent==0)
		return -1;
	if(deep==n){
		if(ent>=a[n])
			return a[n];
		else
			return ent;
	}
	ll tmp=dfs(deep+1,(ent-a[deep])*2);
	if(tmp==-1)
		return -1;
	if(ent-a[deep]>=tmp){
		ans+=tmp+a[deep];
		return tmp+a[deep];
	}
	else{
		ans+=ent;
		return ent;
	}
		
	
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	ull i,tmp=1;
	
	for(i=0;i<=n;++i)
		cin>>a[i];
	ans=a[n];
	if(dfs(0,1)==-1){
		cout<<"-1";
		return 0;
	}
	cout<<ans;
	
	return 0;
}