Alice and Brown loves games. Today, they will play the following game.

In this game, there are two piles initially consisting of X and Y stones, respectively. Alice and Bob alternately perform the following operation, starting from Alice:

* Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile.



The player who becomes unable to perform the operation, loses the game.

Given X and Y, determine the winner of the game, assuming that both players play optimally.

Constraints

* 0 ≤ X, Y ≤ 10^{18}

Input

Input is given from Standard Input in the following format:


X Y


Output

Print the winner: either `Alice` or `Brown`.

Examples

Input

2 1


Output

Brown


Input

5 0


Output

Alice


Input

0 0


Output

Brown


Input

4 8


Output

Alice
x, y = map(int, input().split())
print('Brown' if abs(x - y) <= 1 else 'Alice')

# 総和<=1なら後手勝利（ベース）
# 総和>1かつ差<=1なら、どうしても差>1で相手に手を渡すことになるが、
#   総和>1が確定するので、後手の負けはない
# 総和>1かつ差>1なら、差<=1にできるが、
#   このとき、総和<=1になれば、後手に負けを押し付けられるので、先手勝利

# 総和>1かつ初手で差>1なら、相手に差<=1を押し付けられるので、自分の負けはなく、
# 差<=1を押し付け続けながら、総和<=1にできるのを待てばよいので、先手必勝

# 総和>1かつ初手で差<=1なら、後手は差>1を持てるので、後手必勝
# また、総和<=1のとき、後手必勝だが、差<=1を満たす

# 初手の差で決まる