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 <iostream>

using namespace std;
typedef long long ll;

const int MAXH = 410;
const int MAXW = 410;

int h, w;
string s[MAXH];
bool visited[MAXH][MAXW];
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};

bool check(int x, int y) {
  return 0 <= x && x < w && 0 <= y && y < h;
}

void dfs(int x, int y, ll& white, ll& black) {
  if (s[y][x] == '.') {
    white++;
  } else {
    black++;
  }
  visited[y][x] = true;
  for (int i=0;i<4;i++) {
    int nx = x + dx[i];
    int ny = y + dy[i];
    if (check(nx, ny) && !visited[ny][nx] && s[ny][nx] != s[y][x]) {
      dfs(nx, ny, white, black);
    }
  }
}

void solve() {
  ll cnt = 0;
  for (int i=0;i<h;i++) {
    for (int j=0;j<w;j++) {
      if (!visited[i][j]) {
        ll white = 0, black = 0;
        dfs(j, i, white, black);
        cnt += white * black;
      }
    }
  }
  cout << cnt << endl;
}

int main() {
  cin >> h >> w;
  for (int i=0;i<h;i++) {
    cin >> s[i];
  }
  solve();
}
