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 LL long long
using namespace std;
const int N=1e5+50;
int n,a[N];LL s[N],ans,now;
int main(){
    scanf("%d",&n);
    for(int i=0;i<=n;i++)scanf("%d",&a[i]);
    for(int i=n;i;i--)s[i]=s[i+1]+a[i];
    for(int i=0;i<=n;i++){
        now=i?min(now*2,s[i]):1,ans+=now;
        if(now<a[i])puts("-1"),exit(0);
        now-=a[i];
    }
    cout<<ans;
    return 0;
}