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
#(1,0)と(1,1)が敗戦の最終状態。そこから遷移する状態を考える。
#(2,0),(3,0)は勝利状態。
#(2,1),(2,2)はどう遷移しても上記勝利状態に遷移してしまうので、敗北状態。
#(3,1),(4,0),(4,1),(5,0)...などが上記敗北状態に遷移できる状態なので勝利状態。
#実験を重ねると|X-Y|<=1が敗戦状態ではないか、という仮説が立つ。
#少し考えると、|X-Y|<=1からの遷移は必ず|X'-Y'|>1にならざるを得なくて
#|X-Y|>1からは必ず|X'-Y'|<=1となる遷移が存在することが分かる。
#最終的には|X-Y|<=1の状態は2N手後に(1,0)と(1,1)へと収斂するので、勝利条件が確定する。

X,Y=map(int,input().split())
if abs(X-Y)<=1:
  print("Brown")
else:
  print("Alice")