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;
int row,col;
bool f[444][444];
int vis[444][444];
int B[444*444],W[444*444],T;
void dfs(int x,int y,bool c){
	if(x < 0 || x >= row || y < 0 || y >= col)return;
	if(f[x][y] != c || vis[x][y] != -1)return;
	vis[x][y] = T;
	bool cc = c;
	if((x+y)%2)cc = !cc;
	if(!cc)W[T] ++;
	dfs(x-1, y, c);
	dfs(x+1, y, c);
	dfs(x, y-1, c);
	dfs(x, y+1, c);
}
int main(){
	char s[999];
	scanf("%d%d",&row,&col);
	for(int i=0; i<row; i++){
		scanf("%s",s);
		for(int j=0; j<col; j++)
			f[i][j] = (s[j] == '#');
	}
	for(int i=0; i<row; i++)for(int j=0; j<col; j++)
		if((i+j)%2)f[i][j] = !f[i][j];
	memset(vis,-1,sizeof(vis));
	for(int i=0; i<row; i++)for(int j=0; j<col; j++){
		if(vis[i][j] == -1){
			dfs(i,j,f[i][j]);
			T++;
		}
	}
	long long int res = 0;
	for(int i=0; i<row; i++)for(int j=0; j<col; j++)
		if((i+j)%2)f[i][j] = !f[i][j];
	for(int i=0; i<row; i++)for(int j=0; j<col; j++)if(f[i][j])
		res += W[vis[i][j]];
	cout << res << endl;
	return 0;
}
