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;
using ll=long long;

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);
    ll n;cin>>n;
    ll arr[n+1];
    ll sum=0;
    bool state=0;
    for(int i=0;i<=n;i++){
        cin>>arr[i];sum+=arr[i];
    }
    if(n==0){if(arr[0]==1){cout<<"1\n";return 0;}if(arr[0]==2){cout<<"-1\n";return 0;}}
    ll cnt=1,node=1;
    for(int i=1;i<=n;i++){
        node=min(node*2,sum);
        cnt+=node;
        node-=arr[i];
        sum-=arr[i];
        if(node<0)state=1;
    }
    if(state||arr[0]>0){cout<<"-1\n";return 0;}
    cout<<cnt<<"\n";
}
