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: Accepted)

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
int const MOD = 1000000007;
ll mod_pow(ll x, ll n) {return (!n)?1:(mod_pow((x*x)%MOD,n/2)*((n&1)?x:1))%MOD;}
int madd(int a, int b) {return (a + b) % MOD;}
int msub(int a, int b) {return (a - b + MOD) % MOD;}
int mmul(int a, int b) {return (a * b) % MOD;}
int minv(int a) {return mod_pow(a, MOD-2);}
int mdiv(int a, int b) {return mmul(a, minv(b));}

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

int memo[1010][1010][2];

int solve(int x, int y, int turn) {
    // printf("x = %lld, y = %lld, turn = %lld\n", x, y, turn);
    if(x > y) swap(x, y);
    if(memo[x][y][turn] != -1) return memo[x][y][turn];
    if(x == 0 && y == 0) return memo[x][y][turn] = (turn % 2 ? 0 : 1);
    if(x == 0 && y == 1) return memo[x][y][turn] = (turn % 2 ? 0 : 1);
    if(x == 1 && y == 1) return memo[x][y][turn] = (turn % 2 ? 0 : 1);

    bool ok = false;
    for(int t = 2; t <= x; t += 2) {
        if(turn % 2 && solve(x-t, y+t/2, turn^1)) ok = true;
        else if(turn % 2 == 0 && !solve(x-t, y+t/2, turn^1)) ok = true;
    }
    for(int t = 2; t <= y; t += 2) {
        if(turn % 2 && solve(x+t/2, y-t, turn^1)) ok = true;
        else if(turn % 2 == 0 && !solve(x+t/2, y-t, turn^1)) ok = true;
    }
    if(turn % 2) {
        return memo[x][y][turn] = (ok ? 1 : 0);
    }
    else return memo[x][y][turn] = (ok ? 0 : 1);
}

signed main() {
    int x, y; cin >> x >> y;
    if(x > y) swap(x, y);
    if(y == x || y == x+1) cout << "Brown" << endl;
    else cout << "Alice" << endl;
    

    /*
    memset(memo, -1, sizeof(memo));
    rep(i,0,100) rep(j,i,100) {
        int ans = solve(i, j, 0);
        // if(ans == 0) continue;
        printf("%lld, %lld, ", i, j);
        cout << (ans ? "Brown" : "Alice") << endl;
    }
    */
    return 0;
}