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=1e5+6;
const ll inf=1e14;
int n,a[N];
ll b[N];
int main()
{
    scanf("%d",&n);
    for(int i=0;i<=n;i++) scanf("%d",&a[i]);
    b[0]=1-a[0];
    if(b[0]<0)
    {
        printf("-1\n");return 0;
    }
    for(int i=1;i<=n;i++)
    {
        b[i]=b[i-1]*2-a[i];
        b[i]=min(b[i],inf);
        if(b[i]<0)
        {
            printf("-1\n");return 0;
        }
    }
    ll ans=a[n],sum=a[n];
    for(int i=n-1;i>=0;i--)
    {
        sum=min(sum,b[i])+a[i];
        ans+=sum;
    }
    printf("%lld\n",ans);
}
