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 <algorithm>
#include <vector>
#include <utility>
#include <map>

using namespace std;

long long cnt[500][500];
int ch[500][500][2];
bool used[500][500];
char s[500][500];
int dx[]={0,0,-1,1};
int dy[]={-1,1,0,0};
int h,w;

long long dfs(int x, int y, int px, int py){
  used[x][y]=true;
  ch[x][y][0]=px;
  ch[x][y][1]=py;
  long long sum = 0;
  for(int i=0;i<4;i++){
    if(x+dx[i]<0 || x+dx[i]>=h || y+dy[i]<0 || y+dy[i]>=w)continue;
    if(used[x+dx[i]][y+dy[i]])continue;
    if(s[x+dx[i]][y+dy[i]]!=s[x][y]){
      sum += dfs(x+dx[i],y+dy[i],px,py);
    }
  }
  if(s[x][y]=='#')sum++;
  return sum;
}

int main(){
  long long int i,j,k,ret=0;
  cin >> h >> w;
  for(i=0;i<h;i++)for(j=0;j<w;j++)cin >> s[i][j];
  for(i=0;i<h;i++)for(j=0;j<w;j++){
    if(!used[i][j])cnt[i][j]=dfs(i,j,i,j);
  }
  for(i=0;i<h;i++)for(j=0;j<w;j++){
    if(s[i][j]=='.')ret+=cnt[ch[i][j][0]][ch[i][j][1]];
  }
  cout << ret << endl;
  return 0;
}
