//連結成分中の白黒の数を数えるだけ。 WAになってないとダメ

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};

int main(){
  int n,m;
  cin >> n >> m;
  vector<string> s(n);
  for(auto& x : s) cin >> x;

  vector<vector<bool>> visit(n, vector<bool>(m, false));
  int ans = 0;
  int b = 0, w = 0;

  for(int r=0; r<n; r++)
  for(int c=0; c<m; c++){
    if(visit[r][c] || s[r][c]=='.') continue;
    queue<int> q;
    q.push(r*m + c);
    visit[r][c] = true;
    int cnt_b = 0;
    int cnt_w = 0;

    while(q.size()){
      int p = q.front(); q.pop();
      int y = p/m;
      int x = p%m;
      cnt_b += s[y][x] == 'b';
      cnt_w += s[y][x] == 'w';

      for(int k__=0; k__<4; k__++){
        int new_x = x + dx[k__];
        int new_y = y + dy[k__];
        if(new_x < 0 || new_x >= m || new_y < 0 || new_y >= n) continue;
        if(visit[new_y][new_x] || s[new_y][new_x] == '.') continue;
        visit[new_y][new_x] = true;
        q.push(new_y * m + new_x);
      }
    }

    int d = min(cnt_b, cnt_w);
    ans += d*100;
    b += cnt_b - d;
    w += cnt_w - d;
  }

  ans += min(b,w) * 10;
  ans += max(b,w) - min(b,w);

  cout << ans << endl;
  return 0;
}