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>
#define MAXN 100007
using namespace std;
long long a[MAXN],m[MAXN];
int main()
{
	int n; scanf("%d",&n);
	for(int i=0;i<=n;i++) scanf("%d",&a[i]);
	if(n==0)
	{
		if(a[0]==1) cout<<1;
		else cout<<-1;
		return 0;
	}
	long long rez=a[0]+a[n]; bool fas=false;
	m[0]=1-a[0];
	if(m[0]<=0) fas=true;
	for(int i=1;i<n;i++)
	{
		m[i]=2*m[i-1]-a[i];
		if(m[i]<=0) fas=true;
		if(m[i]>1000000000000000000LL) m[i]=1000000000000000000LL;
		rez+=a[i];
	}
	long long dt=a[n];
	for(int i=n-1;i>=0;i--)
	{
		if(dt>2*m[i]) fas=true;
		if(dt>m[i]) dt=m[i];
		rez+=dt;
		dt+=a[i];
	}
	if(fas) printf("-1");
	else printf("%lld",rez);
}