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 <iostream>
#include <math.h>
#include <algorithm>
#include <string>
using namespace std;
long long m[100001];
long long A[100001];
int main(){
  int N;
  cin>>N;
  long long nord=1;
  long long answer=0;
  long long M=0;
  for(int i=0;i<=N;i++){
    cin>>A[i];
    M+=A[i];
  }
  for(int i=0;i<=N;i++){
    m[i]=M;
    M-=A[i];
  }
  for(int i=0;i<=N+1;i++){
    answer+=nord;
    if(i<=N) nord-=A[i];
    if(nord<0){
      cout<<-1<<endl;
      break;
    }
    if(i<N) nord=min(nord*2,m[i+1]);
    if(i==N+1) cout<<answer<<endl;
  }
  return 0;
}