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<queue>
#include<string>
using namespace std;
typedef long long ll;
ll memo[405][405]={0};
vector<ll> black,white;
string s[405];
int h,w;
void dfs(ll x,ll y,ll k){
  int dx[]={-1,0,1,0},dy[]={0,-1,0,1};
  memo[x][y]=k;
  if(s[x][y]=='#') black[k]++;
  else white[k]++;
  for(int i=0;i<4;i++){
    int nx=x+dx[i],ny=y+dy[i];
    if(nx>=0&&nx<h&&ny>=0&&ny<w){
      if(s[x][y]!=s[nx][ny]){
        if(memo[nx][ny]==0) dfs(nx,ny,k);
      }
    }
  }

}
int main(){
  ll ans=0;
  cin>>h>>w;
  for(int i=0;i<h;i++) cin>>s[i];
  ll cnt=1;
  black.push_back(0);
  white.push_back(0);
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(memo[i][j]==0) {
        black.push_back(0);
        white.push_back(0);
        dfs(i,j,cnt);
        cnt++;
      }
    }
  }
  /*for(int i=0;i<h;i++){
    for(int j=0;j<w;j++) cout<<memo[i][j]<<" ";
    cout<<endl;
  }*/
  for(int i=1;i<black.size();i++) ans+=black[i]*white[i];
  cout<<ans<<endl;
}
