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 rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int a[100010];
ll mn[100010],mx[100010];

int main(){
    int n;
    cin>>n;
    rep(i,n+1)cin>>a[i];
    mn[n]=mx[n]=a[n];
    for(int i=n; i>0; --i){
        mn[i-1]=a[i-1]+(mn[i]+1)/2;
        mx[i-1]=a[i-1]+mx[i];
    }
    if(mn[0]!=1){cout<<-1<<endl; return 0;}
    ll ans=0,now=1;
    rep(i,n+1){
        ans+=now;
        now-=a[i];
        now=min(now*2,mx[i+1]);
    }
    cout<<ans<<endl;
}