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 <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define FORR(i, a, b) for(ll i = (a); i > (b); --i)
#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOREACH(x, a) for(auto &(x) : (a))
#define VECCIN(x)                                                              \
    for(auto &youso_ : (x)) cin >> youso_
#define bitcnt(x) __builtin_popcount(x)
#define lbit(x) __builtin_ffsll(x)
#define rbit(x) __builtin_clzll(x)
#define SZ(x) ((ll)(x).size())
#define fi first
#define se second
#define All(a) (a).begin(), (a).end()
#define rAll(a) (a).rbegin(), (a).rend()

template <typename T = long long> inline T IN() {
    T x;
    cin >> x;
    return (x);
}
inline void CIN() {}
template <class Head, class... Tail>
inline void CIN(Head &&head, Tail &&... tail) {
    cin >> head;
    CIN(move(tail)...);
}
#define CCIN(...)                                                              \
    char __VA_ARGS__;                                                          \
    CIN(__VA_ARGS__)
#define DCIN(...)                                                              \
    double __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define LCIN(...)                                                              \
    ll __VA_ARGS__;                                                            \
    CIN(__VA_ARGS__)
#define SCIN(...)                                                              \
    string __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define Yes(a) cout << (a ? "Yes" : "No") << "\n"
#define YES(a) cout << (a ? "YES" : "NO") << "\n"
#define Printv(v)                                                              \
    {                                                                          \
        FOREACH(x, v) { cout << x << " "; }                                    \
        cout << "\n";                                                          \
    }
template <typename T = string> inline void eputs(T s) {
    cout << s << "\n";
    exit(0);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *)array, (T *)(array + N), val);
}

template <typename T> using PQG = priority_queue<T, vector<T>, greater<T>>;
template <typename T> using PQ = priority_queue<T>;

typedef long long ll;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<ll, ll> PL;
typedef vector<PL> VPL;
typedef vector<bool> VB;
typedef vector<double> VD;
typedef vector<string> VS;

const int INF = 1e9;
// const int MOD = 1e9 + 7;
const int MOD = 998244353;
const ll LINF = 1e18;
// const double PI = atan(1.0) * 4.0;
const ll dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
#define PI 3.141592653589793238

ll dp[51][51];

ll calcdp(ll h, ll w) {
    if(dp[h][w] != -1) return dp[h][w];
    bitset<5000> bit;
    FOR(i, 1, h / 2 + 1) { bit.set(calcdp(h - i * 2, w + i)); }
    FOR(i, 1, w / 2 + 1) { bit.set(calcdp(h + i, w - 2 * i)); }
    ll val = 0;
    while(bit.test(val)) val++;
    return dp[h][w] = val;
}

signed main() {
    LCIN(X, Y);
    // Fill(dp, -1);
    // REP(i, X + 1) REP(j, Y + 1) calcdp(i, j);
    // REP(i, X + 1) REP(j, Y + 1) { cout << dp[i][j] << (j == Y ? "\n" : " "); }
    cout << (abs(X - Y) <= 1 ? "Brown" : "Alice") << "\n";
}