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
//Love and Freedom.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
#define inf 20021225
#define int ll
using namespace std;

char ch[440][440];
int id[440][440],h,w,ans;
bool vis[440][440];
char mp[440][440];
int bl,wh;
int xx[4] = {0,0,-1,1};
int yy[4] = {1,-1,0,0};

void dfs(int x,int y)
{
	if(ch[x][y]=='#')	bl++;
	else	wh++;
	vis[x][y] = 1;
	for(int i=0;i<4;i++)
	{
		int tx = x+xx[i], ty=y+yy[i];
		if(tx<1||ty<1||tx>h||ty>w||vis[tx][ty])	continue;
		if(id[tx][ty] == id[x][y])	dfs(tx,ty);
	}
}

signed main()
{
	scanf("%lld%lld",&h,&w);
	for(int i=1;i<=h;i++)
		scanf("%s",ch[i]+1);
	for(int i=1;i<=h;i++)
		for(int j=1;j<=w;j++)
			mp[i][j] = ((i&1)^(j&1)?'#':'.');
	for(int i=1;i<=h;i++)
		for(int j=1;j<=w;j++)
			id[i][j] = (ch[i][j] == mp[i][j]);
	for(int i=1;i<=h;i++)
		for(int j=1;j<=w;j++)
			if(!vis[i][j])
				bl=wh=0,dfs(i,j),ans+=bl*wh;
	printf("%lld\n",ans);
	return 0;
}