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>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
  int H,W;
  cin >> H >> W;
  string S[H];
  rep(i,H){
    cin >> S[i];
  }
  int a[H][W];
  rep(i,H)rep(j,W){
    a[i][j]=-1;
  }
  int dh[4]={0,1,0,-1};
  int dw[4]={1,0,-1,0};
  int cnt=0;
  ll ans=0;
  queue<pii> q;
  rep(i,H)rep(j,W){
    if(a[i][j]==-1){
      ll w=0;
      ll b=0;
      q.push(pii(i,j));
      cnt++;
      a[i][j]=cnt;
      while(!q.empty()){
        pii n=q.front();
        q.pop();
        if(S[n.first][n.second]=='#') b++;
        else w++;
        rep(k,4){
          int nh=n.first+dh[k];
          int nw=n.second+dw[k];
          if(nh>=0 and nh<H and nw>=0 and nw<W){
            if(S[n.first][n.second]!=S[nh][nw] and a[nh][nw]==-1){
              q.push(pii(nh,nw));
              a[nh][nw]=cnt;              
            }
          }
        }
      }
      ans+=b*w;
    }
  }
  cout << ans << endl;
  return 0;
}
