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;
typedef long long ll;
int h, w;
ll ans, aa, bb;
bool used[410][410];
char maze[410][410];
int dh[4] = {0,1,0,-1};
int dw[4] = {1,0,-1,0};

void dfs(int hh, int ww, char c) {
  if (used[hh][ww]) return;
  used[hh][ww] = 1;
  if (c == '#') aa++;
  else bb++;
  for (int i = 0; i < 4; i++) {
    int th = hh + dh[i], tw = ww + dw[i];
    if (0 <= th && th < h && 0 <= tw && tw < w && maze[th][tw] != c) {
      dfs(th, tw, maze[th][tw]);
    }
  }
}


int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> h >> w;
  for (int i = 0; i < h; i++) cin >> maze[i];
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      if (!used[i][j]) {
        aa = 0, bb = 0;
        dfs(i, j, maze[i][j]);
        ans += aa * bb;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
