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<cstdio>
#define For(i,A,B) for(i=(A);i<=(B);++i)
typedef long long ll;
const int N=100050;
int a[N];
ll s[N];
inline ll Min(ll a,ll b){return a<b?a:b;}
int main(){
	int n,i;
	ll ans=0ll,ret=1ll;
	scanf("%d",&n);
	For(i,0,n)scanf("%d",a+i);
	for(i=n;~i;--i)s[i]=s[i+1]+a[i];
	For(i,0,n){
		if(ret<a[i]){
			puts("-1");
			return 0;
		}
		ans+=ret;
		ret=Min(ret-a[i]<<1,s[i+1]);
	}
	printf("%lld\n",ans);
	return 0;
}