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;
inline void E(){puts("-1");exit(0);}
typedef long long ull;
const int N=1e5+5;
int n,i,a[N];ull l[N],r[N],ans,v[N];
int main(){
	scanf("%d",&n);
	for(i=0;i<=n;++i)scanf("%d",a+i);
	l[0]=r[0]=1;
	for(i=0;i<n;++i){
		if(r[i]<=a[i])E();
		if(l[i]<=a[i])l[i+1]=1;else l[i+1]=l[i]-a[i];
		r[i+1]=r[i]-a[i]<<1;r[i+1]=min(r[i+1],1ll<<50);
	}
	if(l[n]>a[n] || r[n]<a[n])E();
	v[n]=a[n];
	for(i=n-1;i>=0;--i){
		v[i]=min(r[i],v[i+1]+a[i]);
	}
	for(i=0;i<=n;++i)ans+=v[i];
	cout<<ans<<endl;
}