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
import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
    private static boolean debug = false;
    private static boolean elapsed = false;

    private static PrintWriter _out = new PrintWriter(System.out);
    private static PrintWriter _err = new PrintWriter(System.err);

    private void solve(Scanner sc) {
        long X = sc.nextLong();
        long Y = sc.nextLong();

        if (Math.abs(X - Y) <= 1) {
            _out.println("Brown");
        } else {
            _out.println("Alice");
        }
    }
    private static BigInteger C(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        for (long i = r; i > 1; --i) {
            res = res.divide(BigInteger.valueOf(i));
        }
        return res;
    }
    private static BigInteger P(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        return res;
    }
    /*
     * 10^10 > Integer.MAX_VALUE = 2147483647 > 10^9
     * 10^19 > Long.MAX_VALUE = 9223372036854775807L > 10^18
     */
    public static void main(String[] args) {
        long S = System.currentTimeMillis();

        Scanner sc = new Scanner(System.in);
        new Main().solve(sc);
        _out.flush();

        long G = System.currentTimeMillis();
        if (elapsed) {
            _err.println((G - S) + "ms");
        }
        _err.flush();
    }
}