Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

2 1

Sample Output 1

Brown

Alice can do nothing but taking two stones from the pile containing two stones. As a result, the piles consist of zero and two stones, respectively. Then, Brown will take the two stones, and the piles will consist of one and zero stones, respectively. Alice will be unable to perform the operation anymore, which means Brown's victory.

Sample Input 2

5 0

Sample Output 2

Alice

Sample Input 3

0 0

Sample Output 3

Brown

Sample Input 4

4 8

Sample Output 4

Alice

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES

#include <bits/stdc++.h>
using namespace std;
using i32 = int; using i64 = long long int; using f64 = double; using str = string;
template <typename T> using vec = vector<T>;
template <typename T> using heap = priority_queue<T, vec<T>, greater<T>>;
#define times(n, i) for (i32 i = 0; i < (n); ++i)
#define range(a, b, i) for (i32 i = (a); i < (b); ++i)
#define upto(a, b, i) for (i32 i = (a); i <= (b); ++i)
#define downto(a, b, i) for (i32 i = (a); i >= (b); --i)
#define all(xs) (xs).begin(), (xs).end()
#define sortall(xs) sort(all(xs))
#define reverseall(xs) reverse(all(xs))
#define uniqueall(xs) (xs).erase(unique(all(xs)), (xs).end())
#define even(x) (((x) & 1) == 0)
#define odd(x) (((x) & 1) == 1)
#define head(xs) *(xs).begin()
#define last(xs) *(xs).rbegin()
#define append emplace_back
#define findge lower_bound
#define findgt upper_bound
const i64 MOD = 1000000007;
const f64 EPS = 1e-10;

i64 x, y;

i64 g(i64 a, i64 b) {
    i64 t = max(1LL, a / 2 - 2);
    set<i64> s;
    while (2 * t <= a) {
        s.insert(g(a - 2 * t, b + 1));
        t++;
    }
    t = max(1LL, b / 2 - 2);
    while (2 * t <= b) {
        s.insert(g(a + 1, b - 2 * t));
        t++;
    }
    i64 i = 0;
    for (i64 v : s) {
        if (v != i) {
            return i;
        }
        i++;
    }
    return i;
}

i32 main()
{
    cin >> x >> y;
    cout << (g(x, y) == 0 ? "Brown" : "Alice") << endl;
    return 0;
}