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;

typedef long long LL;

const int N=200000;

int n,a[N+9];

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