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;

int H, W;
vector<string> S;
vector<vector<bool>> visited;
vector<int> dy = { 1, -1, 0, 0 };
vector<int> dx = { 0, 0, -1, 1 };
long long black, white;

void dfs(int y, int x) {
  visited.at(y).at(x) = true;
  if (S.at(y).at(x) == '#') black++;
  else white++;
  for (int d = 0; d < 4; d++) {
    int ny = y + dy.at(d);
    int nx = x + dx.at(d);
    if (ny < 0 || ny >= H || nx < 0 || nx >= W) continue;
    if (visited.at(ny).at(nx)) continue;
    if (S.at(ny).at(nx) == S.at(y).at(x)) continue;
    dfs(ny, nx);
  }
}

int main() {
  cin >> H >> W;
  S.resize(H);
  for (int i = 0; i < H; i++) cin >> S.at(i);
  visited = vector<vector<bool>>(H, vector<bool>(W));
  long long ans = 0;
  for (int y = 0; y < H; y++) {
    for (int x = 0; x < W; x++) {
      if (S.at(y).at(x) == '.') continue;
      if (visited.at(y).at(x)) continue;
      black = 0; white = 0;
      dfs(y, x);
      ans += black * white;
    }
  }
  cout << ans << endl;
}