There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

* There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

* 1 \leq H, W \leq 400
* |S_i| = W (1 \leq i \leq H)
* For each i (1 \leq i \leq H), the string S_i consists of characters `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
S_2
:
S_H


Output

Print the answer.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
#include <bits/stdc++.h>
using namespace std;

#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >= 0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)

using ll = long long;
#define INF 1e9

int H, W;
vector<string> S;
vector<vector<bool>> homon;
ll black =0, white = 0;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};

void dfs(int y, int x){
  homon[y][x] = true;
  if(S[y][x] == '#'){
    black++;
  }
  else{
    white++;
  }
  REP(i, 4){
    int nx = x + dx[i], ny = y + dy[i];
    if(0> nx || W<=nx || 0 > ny || H <= ny || homon[ny][nx] || S[y][x] == S[ny][nx]) continue;
    dfs(ny, nx);
  }
}

int main(){ 
  cin >> H >> W;
  S.resize(H), homon.resize(H, vector<bool>(W, false));
  REP(i, H){
    cin >> S[i];
  }
  ll ans = 0;
  REP(i, H){
    REP(j, W){
      if(S[i][j] == '.' || homon[i][j]) continue;
      black = 0, white = 0;
      dfs(i, j);
      ans += black * white;
    }
  }
  cout << ans << endl;
  return 0;
}

