#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

const int N = 30;
string s[N];


string st = "RBG";
int a[N][N]; //1 - row, 2 - col, 3 - row & col

int dp[N][N][N][N];

int solve(int lx, int rx, int ly, int ry){
	if(lx >= rx || ly >= ry)
		return 0;
	if(dp[lx][rx][ly][ry] != -1)
		return dp[lx][rx][ly][ry];

	VI cur;
	FOR(x, lx, rx){
		FOR(y, ly, ry){
			if(a[x][y] == 0)
				continue;
			if(a[x][y] == 1)
				cur.PB(solve(lx, x, ly, ry) ^ solve(x + 1, rx, ly, ry));
			if(a[x][y] == 2)
				cur.PB(solve(lx, rx, ly, y) ^ solve(lx, rx, y + 1, ry));
			if(a[x][y] == 3){
				cur.PB(solve(lx, x, ly, y) ^ solve(x + 1, rx, ly, y) ^ solve(lx, x, y + 1, ry) ^ solve(x + 1, rx, y + 1, ry));
		//		cout << lx << " " << x << " " << ly << " " << y << endl;
			}
		}
	}	
//	cout << lx << " " << rx << " " << ly << " " << ry << endl;
//	cout << SZ(cur) << endl;
	VI cnt(SZ(cur) + 1);
	for(int i : cur)
		if(i < SZ(cnt))
			cnt[i] = 1;
	int mex = 0;
	while(cnt[mex] == 1)
		mex++;
	
	return dp[lx][rx][ly][ry] = mex;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m;
	cin >> n >> m;
	FOR(i, 0, n)
		cin >> s[i];
	if(n == 1 && m == 1){
		cout << "W" << endl;
		return 0;
	}
	
		
	
	int ans = 0;
	FOR(t, 0, 2){
		FOR(i, 0, N)
		FOR(j, 0, N)
			a[i][j] = 0;
		FILL(dp, -1);
		
		
		int Lx = 2, Rx = 0;
		int Ly = 2, Ry = 0;
		FOR(i, 0, n){
			FOR(j, 0, m){
				if((i + j) % 2 != t)
					continue;
				int x = (i + j + t) / 2;//x-th row
				int y = (i - j + t) / 2;//y-th col
				
				Lx = min(Lx, x);
				Rx = max(Rx, x);
				Ly = min(Ly, y);
				Ry = max(Ry, y);
			}
		}
		
		FOR(i, 0, n){
			FOR(j, 0, m){
				if((i + j) % 2 != t)
					continue;
				int x = (i + j + t) / 2;//x-th row
				int y = (i - j + t) / 2;//y-th col
				
				x -= Lx;
				y -= Ly;
				FOR(k, 0, 3)
					if(s[i][j] == st[k])
						a[x][y] = k + 1; 
			}
		}
		
		
		//cout << Rx - Lx << " " << Ry - Ly << endl;
		ans ^= solve(0, Rx - Lx + 1, 0, Ry - Ly + 1);
		//cout << ans << endl;
	}
	cout << ((ans == 0) ? "L" : "W") << endl;
	
	return 0;
}