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
d,l = [list(map(int, s.split())) for s in open(0)]
d = d[0]

ans = 0
v_max = [1]
for i in range(d+1):
    if v_max[i] - l[i] < 0:
        ans = -1
        break
    v_max.append((v_max[i] - l[i])*2)
else:
    _max = 0
    for i in range(d, -1, -1):
        _max = min(_max + l[i], v_max[i])
        ans += _max
print(ans)