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 h,w;
long long ans=0;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
bool vis[405][405];
char a[405][405];
pair<int,int> dfs(int x,int y)
{
    pair<int,int>v;
    if(a[x][y]=='.') v.second=1,v.first=0;
    else v.second=0,v.first=1;
    for(int i=0;i<4;i++)
    {
        int xx=x+dx[i];
        int yy=y+dy[i];
        if(xx<1||xx>h||yy<1||yy>w||vis[xx][yy]||a[xx][yy]==a[x][y]) continue;
        vis[xx][yy]=true;
        pair<int,int>t=dfs(xx,yy);
        v.first+=t.first;
        v.second+=t.second;
    }
    return v;
}
int main()
{
    scanf("%d%d",&h,&w);
    for(int i=1;i<=h;i++) scanf("%s",a[i]+1);
    for(int i=1;i<=h;i++)
        for(int j=1;j<=w;j++)
        {
            if(vis[i][j]) continue;
            vis[i][j]=true;
            pair<int,int>v=dfs(i,j);
            ans+=(long long)v.first*v.second;
        }
    printf("%lld\n",ans);
}