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>

#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const int MAX = 410;

int h, w;
vector<string> fi;

long long numb = 0, numw = 0;
bool seen[MAX][MAX];
void dfs(int x, int y) {
  seen[x][y] = true;
  if (fi[x][y] == '#')
    ++numb;
  else
    ++numw;

  rep(dir, 4) {
    int nx = x + dx[dir];
    int ny = y + dy[dir];
    if (nx < 0 || nx >= h || ny < 0 || ny >= w || fi[x][y] == fi[nx][ny] ||
        seen[nx][ny])
      continue;
    dfs(nx, ny);
  }
}

int main() {
  cin >> h >> w;
  fi.resize(h);
  rep(i, h) cin >> fi[i];

  long long res = 0;
  rep(x, h) {
    rep(y, w) {
      if (fi[x][y] == '.' || seen[x][y]) continue;
      numb = numw = 0;
      dfs(x, y);
      res += numb * numw;
    }
  }
  cout << res << endl;
}
