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)

/*input
2 1
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#ifdef leowang
#define debug(...) do{\
	fprintf(stderr,"%s - %d : (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
	_DO(__VA_ARGS__);\
}while(0)
template<typename I> void _DO(I&&x){cerr<<x<<endl;}
template<typename I,typename...T> void _DO(I&&x,T&&...tail){cerr<<x<<", ";_DO(tail...);}
#else
#define debug(...)
#endif
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
	out<<'('<<P.F<<','<<P.S<<')';
	return out;
}
//}}}
const ll maxn=300005;
const ll maxlg=__lg(maxn)+2;
const ll INF64=8000000000000000000LL;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const double PI=acos(-1);
//const ll p=880301;
//const ll P=31;

ll mypow(ll a,ll b){
	ll res=1LL;
	while(b){
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}



int main()
{
	//IOS;
	ll x,y;
	cin>>x>>y;
	if(abs(x-y)<=1) cout<<"Brown\n";
	else cout<<"Alice\n";

	return 0;
}