#include <iostream>
#include <vector>
#include <string>
#include <deque>
#include <algorithm>
using namespace std;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int main() {
// Tezkor kiritish va chiqarish
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int H, W;
cin >> H >> W;
vector<string> meadow(H);
for (int i = 0; i < H; i++) {
cin >> meadow[i];
}
// Masofalarni saqlash uchun massiv (-1 hali ko'rilmagan degani)
vector<vector<int>> dist(H, vector<int>(W, -1));
deque<pair<int, int>> dq;
// Boshlang'ich nuqtani belgilaymiz
dist[0][0] = 1;
dq.push_back({0, 0});
int max_depth = 1;
while (!dq.empty()) {
pair<int, int> curr = dq.front();
dq.pop_front();
int r = curr.first;
int c = curr.second;
max_depth = max(max_depth, dist[r][c]);
for (int i = 0; i < 4; i++) {
int nr = r + dx[i];
int nc = c + dy[i];
// Maydon ichida ekanligini va qor emasligini tekshiramiz
if (nr >= 0 && nr < H && nc >= 0 && nc < W && meadow[nr][nc] != '.' && dist[nr][nc] == -1) {
if (meadow[nr][nc] == meadow[r][c]) {
// Bir xil turdagi iz - masofa 0
dist[nr][nc] = dist[r][c];
dq.push_front({nr, nc});
} else {
// Har xil turdagi iz - masofa 1 (yangi hayvon)
dist[nr][nc] = dist[r][c] + 1;
dq.push_back({nr, nc});
}
}
}
}
cout << max_depth << endl;
return 0;
}
