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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int H;
int W;
vector<string> S;
int cnt;

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int d[401][401];

int b = 0;
int dfs(int y,int x,int p){
  if( x < 0 || y < 0 || x >= W || y >= H) return 0;
  if( S[y][x] == p ) return 0;
  if( d[y][x]++ ) return 0;
  cnt++;
  b += S[y][x] == '#';
  for(int i = 0 ; i < 4 ; i++){
    dfs(y + dy[i], x + dx[i], S[y][x]);
  }
  return 0;
}
void solve(long long H, long long W, vector<string> S) {
  ::H = H;
  ::W = W;
  ::S = S;
  long long cnt2 = 0;
  rep(i, H) rep(j, W){
    if( !d[i][j] && S[i][j] == '#'){
      cnt = 0;
      b = 0;
      dfs(i, j, 0);
      cnt2 += 1ll * b * (cnt-b);
    }
  }
  cout << cnt2 << endl;
}

int main() {

  long long H;
  scanf("%lld", &H);
  long long W;
  scanf("%lld", &W);
  vector<string> S(H);
  for (int i = 0; i < H; i++) {
    cin >> S[i];
  }
  solve(H, W, S);
  return 0;
}
