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 fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long LL;
typedef pair<int,int> pr;

const int maxn=1e5+5;

int n,a[maxn];
LL s[maxn];

int main()
{
	scanf("%d",&n);
	fo(i,0,n) scanf("%d",&a[i]);
	fd(i,n,0) s[i]=s[i+1]+a[i];
	
	LL now=1, ans=0;
	fo(i,0,n)
	{
		if (i) now<<=1;
		now=min(now,s[i]);
		ans+=now;
		now-=a[i];
		if (now<0) {puts("-1"); return 0;}
	}
	
	printf("%lld\n",ans);
}