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 <cstdio>
#include <algorithm>
using namespace std;
int n,m;
char p[405][405];
int bel[405][405];
bool vis[405][405];
int b,w;
long long ans;
void dfs(int x,int y,int cur)
{
	if (x<=0||x>n||y<=0||y>m||vis[x][y]||bel[x][y]!=cur)
		return;
	vis[x][y]=1;
	if (p[x][y]=='#')
		b++;
	else
		w++;
	dfs(x-1,y,cur);
	dfs(x,y-1,cur);
	dfs(x+1,y,cur);
	dfs(x,y+1,cur);
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%s",p[i]+1);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			bel[i][j]=(p[i][j]=='.')^((i^j)&1);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (!vis[i][j])
			{
				b=w=0;
				dfs(i,j,bel[i][j]);
				ans+=1ll*b*w;
			}
	printf("%lld\n",ans);
	return 0;
}