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;
using P = pair<int,int>;
using ull= unsigned long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  ll n;
  cin >>n;
  vector<ll> a(n+1),b(n+1),a2(n+1);
  ll nl=0;//葉の数
  rep(i,n+1) {
    cin >>a[i];
    nl+=a[i];
  }
  if(a[0]>1){
    cout<<-1<<endl;
    return 0;
  }
  
  b[0]=1-a[0];
  ll ans=1;
  nl-=a[0];
  rep(i,n){
    nl-=a[i+1];//a[i+2]以降の葉の数．
    if(a[i+1]>b[i]*2){
      cout<<-1<<endl;
      return 0;
    }
 
    b[i+1]=min(nl, 2*b[i]-a[i+1]);
    ans+=b[i+1]+a[i+1];
  }
  
  cout<<ans<<endl;
  return 0;
  
  
  
  
}
