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(x) for x in input().split()]

L = [0] * N + [A[N]]    # 下限
U = [0] * N + [A[N]]    # 上限
for i in range(N - 1, -1, -1):
    L[i] = (L[i + 1] + 1) // 2 + A[i]
    U[i] = U[i + 1] + A[i]

if 1 < L[0]:
    print(-1)
    sys.exit()

U[0] = 1
for i in range(1, N + 1):
    U[i] = min(2 * (U[i - 1] - A[i - 1]), U[i])   # 深さiの頂点数

print(sum(U))