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 <cstdint>
#include <vector>
#include <utility>

using vec64 = std::vector<int64_t>;

int64_t solve(int n,const vec64 &as) {
  vec64 maxs(n+1);
  int64_t min=0,max=0;
  for ( int i=n; i>=0; i-- ) {
    auto e=as[i];
    min=(min+1)/2+e;
    maxs[i]=max+=e;
  }
  if ( min>1||max<1 ) {
    return -1;
  }
  int64_t t=1,ans=1;
  for ( int i=1; i<=n; i++ ) {
    ans += t = std::min((t-as[i-1])*2,maxs[i]);
  }
  return ans;
}

int main() {
  int n;
  std::cin >> n;

  vec64 as(n+1);
  for ( auto &e: as ) {
    std::cin >> e;
  }

  std::cout << solve(n,as) << std::endl;
}