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 <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define ll long long

ll ans;
int n, m;
char s[666][666];
bool vis[666][666];

int head, tail;
int q[666666][2];
int w, b;

void work(int x, int y, char a){
	if(!vis[x][y] && s[x][y] != a)
		q[++tail][0] = x, q[tail][1] = y, vis[x][y] = 1;
}

void bfs(int a, int c){
	w = b = head = tail = 0;
	q[++tail][0] = a; q[tail][1] = c;
	vis[a][c] = true;
	while(head < tail){
		int x = q[++head][0], y;
		y = q[head][1];
		if(s[x][y] == '#') b++; else w++;
		work(x + 1, y, s[x][y]), work(x - 1, y, s[x][y]), work(x, y + 1, s[x][y]), work(x, y - 1, s[x][y]);
	}
	ans += 1ll * w * b;
}

int main(){
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; i++) vis[i][0] = vis[i][m + 1] = 1;
	for(int i = 1; i <= m; i++) vis[0][i] = vis[n + 1][i] = 1;
	for(int i = 1; i <= n; i++)
		scanf("%s", s[i] + 1);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
			if(!vis[i][j]) bfs(i, j);
	printf("%lld\n", ans);
	return 0;
}