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 <queue>
#define ll long long
using namespace std;
ll res=0;
int n,m;
char c[410][410];
bool visit[410][410]={0};
int d1[4]={0,0,1,-1};
int d2[4]={1,-1,0,0};
inline bool ok(int x,int y)
{
	return x>=0 && y>=0 && x<n && y<m && !visit[x][y];
}
void bfs(int xt,int yt)
{
    queue<int> q;
    int t=0,tt=0;
	q.push(xt);
	q.push(yt);
	visit[xt][yt]=1;
	while (!q.empty())
	{
		int x=q.front(); q.pop();
		int y=q.front(); q.pop();
		//cout <<x<<" "<<y<<"\n";
		(c[x][y]=='.' ? t : tt)++;
		for (int i=0;i<4;i++)
		{
			int xi=x+d1[i];
			int yi=y+d2[i];
			if (ok(xi,yi) && c[xi][yi]!=c[x][y])
			    q.push(xi),q.push(yi),visit[xi][yi]=1;
		}
	}
	//cout <<"t: "<<t<<"  tt: "<<tt<<"  res(before): "<<res<<"  res(after): "<<res+(ll)t*(ll)tt<<"\n";
	res+=(ll)t*(ll)tt;
}
int main()
{
	cin >>n>>m;
	for (int i=0;i<n;i++)
	    for (int j=0;j<m;j++)
	        cin >>c[i][j];
    for (int i=0;i<n;i++)
	    for (int j=0;j<m;j++)
		    if (!visit[i][j]) bfs(i,j);
    cout <<res;
    return 0;
}