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;
#define _GLIBCXX_DEBUG
int64_t MOD=1000000007;

int main() {
  int N;
  cin>>N;
  bool error=false;
  int64_t ans=0;
  vector<int64_t> S(N+1);
  vector<int64_t> T(N+1);
  for(int i=0;i<=N;i++)cin>>S.at(i);
  T.at(N)=S.at(N);
  for(int i=N-1;i>0;i--) T.at(i)=T.at(i+1)+S.at(i);
  T.at(0)=1;
  for(int i=1;i<=N;i++){
    T.at(i)=min(T.at(i),2*T.at(i-1)-2*S.at(i-1));
    if(T.at(i)<1) error=true;
  }
  if(error||S.at(N)!=T.at(N)) cout<<-1;
  else{
    for(int i=0;i<=N;i++){
      ans+=T.at(i);
    }
    cout<<ans<<endl;
  }
 
}
