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<bits/stdc++.h>
using namespace std;
int main(){
  int h,w;
  cin>>h>>w;
  vector<string> s(h);
  for(int i=0;i<h;i++)
    cin>>s.at(i);
  for(int i=0;i<h;i++)
    for(int j=0;j<w;j++)
      if((i+j)%2==0)
        s.at(i).at(j)='.'+'#'-s.at(i).at(j);
  vector<int> ki={0,0,-1,1};
  vector<int> kj={-1,1,0,0};
  int64_t c=0;
  vector<vector<bool>> v(h,vector<bool>(w,0));
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(v.at(i).at(j))
        continue;
      vector<int> p(2,0);
      queue<vector<int>> q;
      q.push({i,j});
      while(!q.empty()){
        int xi=q.front().at(0);
        int xj=q.front().at(1);
        q.pop();
        if(v.at(xi).at(xj))
          continue;
        v.at(xi).at(xj)=1;
        p.at((xi+xj)%2)++;
        for(int m=0;m<(int)ki.size();m++){
          int mi=xi+ki.at(m);
          int mj=xj+kj.at(m);
          if(mi>=0&&mi<h&&mj>=0&&mj<w&&s[mi][mj]==s[xi][xj])
            q.push({mi,mj});
        }
      }
      c+=(int64_t)p.at(0)*p.at(1);
    }
  }
  cout<<c<<endl;
}