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
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.Queue;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int h = Integer.parseInt(sa[0]);
		int w = Integer.parseInt(sa[1]);
		char[][] s = new char[h][w];
		for (int i = 0; i < h; i++) {
			s[i] = br.readLine().toCharArray();
		}
		br.close();

		int[] dx = {0, 0, -1, 1};
		int[] dy = {-1, 1, 0, 0};
		boolean[][] visit = new boolean[h][w];
		long ans = 0;
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				if (visit[i][j]) {
					continue;
				}
				long kuro = 0;
				long siro = 0;
				Queue<Integer> que = new ArrayDeque<Integer>();
				que.add(i * 1000 + j);
				visit[i][j] = true;
				while (!que.isEmpty()) {
					int cur = que.poll();
					int x = cur / 1000;
					int y = cur % 1000;
					if (s[x][y] == '#') {
						kuro++;
					} else {
						siro++;
					}
					for (int a = 0; a < 4; a++) {
						int xx = x + dx[a];
						int yy = y + dy[a];
						if (0 <= xx && xx < h && 0 <= yy && yy < w
								&& !visit[xx][yy] && s[x][y] != s[xx][yy]) {
							que.add(xx * 1000 + yy);
							visit[xx][yy] = true;
						}
					}
				}
				ans += kuro * siro;
			}
		}
		System.out.println(ans);
	}
}
