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; 
ll cnt = 0;
const ll MOD = 1e9 + 7;
int H,W;vector<vector<bool>> visited;
vector<string> S; ll b,w;

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

void dfs(int s,int t){
  visited[s][t] = true;
  if( S[s][t] == '#'){ b++;} else{ w++;}
  for(int k = 0; k < 4; k++){
    int p = s+dx[k]; int q = t+dy[k];
    if( p < 0 || p >= H || q < 0 || q >= W){ continue;}
    if(S[p][q] != S[s][t]){
     if(visited[p][q]){ //こうなることはないと思うが...
                       continue;}
      dfs(p,q);}
  }
  return ;
}
      
       
    
  
int main(){
  cin >> H >> W; S.resize(H);
  visited = vector<vector<bool>> (H,vector<bool>(W,false));
  for(int i = 0; i < H; i++){ cin >> S[i];}
  
  for(int i = 0; i < H; i++){
    for(int j = 0; j < W; j++){
      if(visited[i][j]){continue;}
      b = 0; w = 0;
      dfs(i,j);  cnt += b*w; 
    }
  }
  
      
 cout << cnt << endl; return 0;}