#include <bits/stdc++.h>
using namespace std;
#define MX 4001
typedef long long LL;
typedef pair<int, int> PR;
typedef tuple<int, int, int> TP;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 5;
const LL MAXV = (LL)1e18;
using vi = vector<int>;
using vii = vector<vi>;
int H, W;
string md[MX]; //snow meadow
//idea: the footprints (not all) of an animal can be covered by all its successors' footprints.
//Also can be solved using Disjoint Sets instead of BFS/DFS?
int solve() {
cin >> H >> W;
//read the snow meadow info
for (int i = 0; i < H; i++)
cin >> md[i];
queue<PR> Q[2]; //two queues for BFS responding to R and F
int ans = 0;
if (md[0][0] == '.')
return ans;
vector<char> animal(1, md[0][0]);
animal.push_back(md[0][0] == 'F' ? 'R' : 'F'); //alternate animal's footprints to discover
int cur = 0; //an index of animial to reveal the current animal in backward order
Q[cur].push({0, 0});
md[0][0] = 'X'; //visited
while (!Q[cur].empty()) {
ans++;
//BFS to find the CC for current animal's footprints
while (!Q[cur].empty()) {
int r = Q[cur].front().first, c = Q[cur].front().second;
Q[cur].pop();
int dr = 0, dc = 1;
for (int k = 0; k < 4; k++) { //4 directions
int nr = r + dr, nc = c + dc;
if (nr >= 0 && nr < H && nc >= 0 && nc < W && md[nr][nc] != '.' && md[nr][nc] != 'X') {
if (md[nr][nc] == animal[cur]) { //current animial footprint
Q[cur].push({nr, nc});
} else { //immediately preceding animial footprint
Q[cur^1].push({nr, nc}); //different use
// Or you can do: Q[1-cur].emplace_back(nr, nc);
}
md[nr][nc] = 'X'; //visited
}
//the following 2 statements for iterating thru 4 directions
// you can use: int dr[] = {0, 1, 0, -1}, dc[] = {1, 0, -1, 0};
swap(dr, dc);
dc = -dc;
}
}
cur ^= 1; //for immediately preceding animal. or cur = 1 - cur;
}
return ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin.exceptions(cin.failbit);
cout << solve() << endl;
return 0;
}
