Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define mx 100005
using namespace std;
typedef pair<int, int> P;
long long ans;

int h, w, wh[405][405], dx[] = {0,0,-1,1}, dy[] = {1,-1,0,0};
char s[405];

int main(){
  scanf("%d %d\n", &h, &w);
  for(int i = 0; i < h; i++){
    scanf("%s", s);
    for(int j = 0; j < w; j++) wh[i][j] = s[j] == '.';
  }
  for(int i = 0; i < h; i++){
    for(int j = 0; j < w; j++){
      if(wh[i][j] == 2) continue;
      stack<P> st;
      long long nm[2] = {};
      st.push(P(i, j));
      while(!st.empty()){
        P it = st.top(); st.pop();
        for(int k = 0; k < 4; k++){
          int x = it.first + dx[k], y = it.second + dy[k];
          if(0 <= x && x < h && 0 <= y && y < w && wh[x][y] + wh[it.first][it.second] == 1){
            st.push(P(x,y));
          }
        }
        nm[wh[it.first][it.second]]++;
        wh[it.first][it.second] = 2;
      }
      ans += nm[0]*nm[1];
    }
  }
  printf("%lld\n", ans);
}