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;
int main(){
  ll h,w;cin>>h>>w;
  string s[h];
  for(ll i=0;i<h;i++)cin>>s[i];
  bool used[h][w];
  for(ll i=0;i<h;i++)for(ll j=0;j<w;j++)used[i][j]=false;
  ll ans=0;
  ll dx[4]={-1,1,0,0};
  ll dy[4]={0,0,-1,1};
  for(ll i=0;i<h;i++){
    for(ll j=0;j<w;j++){
      if(s[i][j]=='.'||used[i][j]==true)continue;
      ll c=0;
      ll d=0;
      queue<pair<ll,ll>> q;
      q.push(make_pair(i,j));
      used[i][j]=true;
      while(q.size()>0){
        pair<ll,ll> p=q.front();q.pop();
        ll pf=p.first,ps=p.second;
        //cout<<pf<<" "<<ps<<endl;
        if(s[pf][ps]=='#')c++;
        else d++;
        for(ll k=0;k<4;k++){
          ll in=pf+dy[k],jn=ps+dx[k];
          if(in<0||in>=h||jn<0||jn>=w||used[in][jn]==true)continue;
          if(s[pf][ps]==s[in][jn])continue;
          q.push(make_pair(in,jn));
          used[in][jn]=true;
        }
      }
      ans+=c*d;
      //cout<<c<<" "<<d<<endl;
    }
  }
  cout<<ans<<endl;
}
          
