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 int unsigned long long

using namespace std;
int  n,ans=0;
int  a[100005],b[100005],c[100005];
signed  main()
{
    cin>>n;
    n++;
    for (int i=1; i<=n; i++)
        cin>>a[i];

    b[1]=1,c[n]=a[n];
    for (int i=2; i<=n; i++)
        b[i]=(b[i-1]-a[i-1])*2;
    for (int i=n-1; i>=1; i--)
        c[i]=c[i+1]+a[i];
    for (int i=1; i<=n; i++)
        ans+=min(b[i],c[i]);
    for (int i=1; i<=n; i++)
    {
        if (a[i]>b[i])
            return cout<<-1<<endl,0;
    }
    cout<<ans<<endl;

    return 0;
}
