// Source: https://usaco.guide/general/io
#include <bits/stdc++.h>
using namespace std;
const int mx = 4005;
char a[mx][mx];
deque<pair<int,int>>ok;
int dist[mx][mx];
//0/1 dfs, i do not understand the question lol. If both R and F, are in the graph obviously, the minimum amount of animals is 2.
//no ?
// if there is an appearance of 1, then its just 1.
//otherwise 0.
//How could they be more?
//THe above thought is debunked, upon looking at setups impossible.
//I was just thinking of the cases where the minimum was indeed 2, but that is not always the case.1
//if its the same track we want to visit it and clear it all before going to differing tracks.
int h,w;
bool ontrack(int x, int y ){
return (x>=1)&&(y>=1)&&(x<=h)&&(y<=w)&&(a[x][y]!='.');
}
int main() {
cin>>h>>w;
for(int i=1;i<=h;i++){
string s;
cin>>s;
for(int j=1;j<=w;j++){
char g = s[j-1];
a[i][j]=g;
dist[i][j]=0;
}
}
int ans = INT_MIN;
ok.push_front({1,1});
dist[1][1]=1;
while(!ok.empty()){
int x = ok.front().first;
int y = ok.front().second;
ok.pop_front();
ans = max(ans, dist[x][y]);
//consider up down left right
for(int i = -1;i<=1;i+=2){
int nx = x + i;
if(ontrack(nx,y)&&dist[nx][y]==0){
//depth[x][y]==0 means not visited.
//push the same tracks to priority.
//to commit some sort of flood fill.
if(a[x][y]==a[nx][y]){
//same track/
ok.push_front({nx,y});
dist[nx][y]=dist[x][y];
}else{
ok.push_back({nx,y});
dist[nx][y]=dist[x][y]+1;
}
}
}
for(int i=-1;i<=1;i+=2){
int ny=y+i;
if(ontrack(x,ny)&&dist[x][ny]==0){
//dist[x][y]==0 means not visited.
if(a[x][y]==a[x][ny]){
//same track/'
ok.push_front({x,ny});
dist[x][ny]=dist[x][y];
}else{
ok.push_back({x,ny});
//wait until all the last animals track connected components, are done running.
//then this is as if the next layer of a bipartite tree.
dist[x][ny]=dist[x][y]+1;
}
}
}
}
cout<<ans<<"\n";
}
