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 int64_t ll;

int main(){
  ll n;
  cin >> n;
  ll a[n+1],b[n+1],c[n+1];
  for(int i=0;i<=n;i++){
    cin >> a[i];
  }
  if(a[n]==0){cout << -1 << endl;return 0;}
  b[0]=1;c[n]=a[n];
  for(int i=n;i>0;i--){
    c[i-1]=c[i]+a[i-1];
  }
  for(int i=0;i<n;i++){
    if(a[i]>b[i]){cout << -1 << endl;return 0;}
    b[i]-=a[i];
    b[i+1]=min(c[i],2*b[i]);
  }
  if(a[n]>b[n]){cout << -1 << endl;return 0;}
  b[n]-=a[n];
  ll ans=0;
  for(int i=n;i>0;i--){
    c[i]=min(b[i]+a[i],c[i]);
    ans+=min(b[i]+a[i],c[i]);
  }
  cout << ans+1 << endl;
}
