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 unsigned long long ll;

const int maxn= 1e5+5;
ll a[maxn];
ll n,ans;
ll b[maxn];
ll c[maxn];

int main(){

    cin>>n;
    n++;

    for(int i=1;i<=n;i++) cin>>a[i];
    ans=0;
    b[1]=1;
    c[n]=a[n];
    for(int i=2;i<=n;i++){
        b[i] = (b[i-1]-a[i-1])<<1;
    }
    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]);
    int flag=0;
    for(int i=1;i<=n;i++){
        if(a[i]>b[i]){
            flag=1;
            break;
        }
    }
    if(flag) cout<<-1<<endl;
    else cout<<ans<<endl;
    return 0;
}
