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 <iostream>
    #include <math.h>
    using namespace std;
    int main(void){
        int N;
        cin>>N;
        long long int A[N+1];
        long long int count=0;
        for(int i=0;i<N+1;i++){
            cin>>A[i];
            count+=A[i];
        }
        long long int ans=0;
        long long int node=1;
        for(int i=0;i<N+1;i++){
            ans+=node;
            node=(node-A[i])*2;
            count-=A[i];
            node=min(node,count);
            if(node<0){
                cout<<"-1"<<endl;
                return 0;
            }
        }
        cout<<ans<<endl;
    }