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>
#include <vector>
#include <string>
using namespace std;

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

int H, W;
vector<string> v;

long long numb = 0, numw = 0;
bool seen[MAX][MAX];

void dfs(int x, int y) {
  seen[x][y] = true;
  if(v[x][y] == '#') numb++;
  else numw++;

  for(int i = 0; i < 4; i++) {
    int nx = x + dx[i];
    int ny = y + dy[i];
    if(nx < 0 || nx >= H || ny < 0 || ny >= W) continue;
    if(v[x][y] == v[nx][ny]) continue;
    if(seen[nx][ny]) continue;
    dfs(nx, ny);
  }
}

int main() {
  cin >> H >> W;
  v.resize(H);
  for(int i = 0; i < H; i++) cin >> v[i];

  long long ans = 0;

  for(int x = 0; x < H; x++) {
    for(int y = 0; y < W; y++) {
      if(v[x][y] == '.') continue;
      if(seen[x][y]) continue;
      numb = numw = 0;
      dfs(x, y);
      ans += numb * numw;
    }
  }
  cout << ans << endl;
  return 0;
}
