Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#define MAX_H 400
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
string s[MAX_H];
bool used[MAX_H][MAX_H]={};
ll h,w;
void dfs(ll x,ll y,ll& cnt,ll& bc){
  if(used[x][y]){
    return;
  }
  used[x][y]=true;
  cnt++;
  if(s[x][y]=='#'){
    bc++;
  }
  ll vx[]={1,0,-1,0},vy[]={0,1,0,-1};
  ll nx,ny;
  for(int i=0;i<4;i++){
    nx = x+vx[i];
    ny = y+vy[i];
    if(nx>=0&&ny>=0&&nx<h&&ny<w&&s[x][y]!=s[nx][ny]){
      dfs(nx,ny,cnt,bc);
    }
  }
}
signed main(){
  ll ans=0,tmp,bc;
  cin >> h >> w;
  for(int i=0;i<h;i++){
    cin >> s[i];
  }
  for(ll i=0;i<h;i++){
    for(ll j=0;j<w;j++){
      if(!used[i][j]){
        tmp=0;
        bc=0;
        dfs(i,j,tmp,bc);
        ans += bc*(tmp-bc);
      }
    }
  }
  cout << ans << endl;
}