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
import sys
N = int(input())
A = [int(c) for c in input().split()]
B = [0]*(N+1)
B[0] = 1
if A[0]>=1:
  if N==0 and A[0]==1:
    print(1)
  else:
    print(-1)
  sys.exit()
for i in range(1,N+1):
  B[i] = B[i-1]*2-A[i]
  if B[i]<0:
    print(-1)
    sys.exit()
ans = sum(A)
num = 0
for i in range(N,0,-1):
  num = min(num+A[i],B[i-1])
  ans += num
print(ans)