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 <cstdio>
#include <algorithm>

int n;
int A[100001];
long long int S[100002];

int main(){
  scanf("%d", &n);
  long long int sum = 0;
  for(int i = 0; i <= n; ++i) scanf("%d", A+i), sum += A[i];
  for(int i = n; i >= 0; --i){
    S[i] = S[i+1] + A[i];
  }

  long long int ans = 0;
  long long int l = 1;
  int i;
  for(i = 0; i <= n; ++i){
    if(l < A[i]) break;
    ans += l;
    l -= A[i];
    l = std::min(2*l, S[i+1]);
  }

  if(i <= n) puts("-1");
  else printf("%lld\n", ans);

  return 0;
}
