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;
#define int long long
vector<int> xx={0,-1,0,1},yy={1,0,-1,0};
int H,W;
void a(vector<string> &A,vector<vector<bool>> &B,vector<vector<bool>> &C,int x,int y,int &b,int &w){
  if(A[x][y]=='#')
    b++;
  else
    w++;
  C[x][y]=false;
  for(int i=0;i<4;i++){
    int xxx=x+xx[i],yyy=y+yy[i];
    if(0<=xxx && xxx<H && 0<=yyy && yyy<=W && B[x][y]==B[xxx][yyy] && C[xxx][yyy]){
      a(A,B,C,x+xx[i],y+yy[i],b,w);
    }
  }
}
signed main(){
  cin>>H>>W;
  vector<string> A(H);
  for(string &S:A)
    cin>>S;
  vector<vector<bool>> B(H,vector<bool>(W,true)),C(H,vector<bool>(W,true));
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      if((A[i][j]=='.' && (i+j)%2==0) || (A[i][j]=='#' && (i+j)%2==1))
        B[i][j]=false;
  int k=0;
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      if(C[i][j]){
        int b=0,w=0;
        a(A,B,C,i,j,b,w);
        k+=b*w;
      }
  cout<<k<<endl;
}