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;
#define int long long
const int N=100005;
int n,a[N],v[N];
signed main(){
	scanf("%lld",&n);
	for (int i=0;i<=n;i++)scanf("%lld",&a[i]);
	if (n==0){
		if (a[0]==1){
			puts("1");
			return 0;
		}
		else {
			puts("-1");
			return 0;
		}
	}
	if (a[0]){
		puts("-1");
		return 0;
	}
	int Max=1,s=0;
	for (int i=1;i<=n;i++)
		if (a[i]>Max*2){
			puts("-1");
			return 0;
		}
		else v[i-1]=Max,Max=min(1000000000000000ll,Max*2-a[i]);
	int ans=0;
	for (int i=0;i<=n;i++)ans+=a[i];
	for (int i=n-1;i>=0;i--)v[i]=min(v[i],v[i+1]+a[i+1]),ans+=v[i];
	printf("%lld\n",ans+s);
}
