Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
import java.awt.*;
import java.awt.geom.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        int h=sc.nextInt();
        int w=sc.nextInt();
        char[][] f=new char[h][w];
        for (int i = 0; i < h; i++) {
            String t=sc.next();
            for (int j = 0; j < w; j++) {
                f[i][j]=t.charAt(j);
            }
        }
        long ans=0;
        int[] dx={-1,0,1,0};
        int[] dy={0,-1,0,1};
        int[][] node=new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                if(node[i][j]==0) {
                    Queue<Point> q = new ArrayDeque<>();
                    q.add(new Point(i, j));
                    node[i][j]++;
            //    out.println("s:" + q.peek());
                    long tempW = 0, tempB = 0;
                    while (!q.isEmpty()) {
                        Point p = q.poll();
                     //   out.println(p);
                        if (f[p.x][p.y] == '.') tempW++;
                        else tempB++;
                        for (int k = 0; k < 4; k++) {
                            int x = p.x + dx[k];
                            int y = p.y + dy[k];
                            if (0 <= x && x < h && 0 <= y && y < w) {
                                if (node[x][y] == 0 && f[p.x][p.y] != f[x][y]) {
                                    q.add(new Point(x, y));
                                    node[x][y]++;
                                }
                            }
                        }
                    }
                 //   out.println("GraphSize=" + (tempB + tempW));
                 //   out.println("fin");
                    ans += tempB * tempW;
                }
            }
        }
        out.println(ans);
/*
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                out.print(node[i][j]);
            }
            out.println();
        }*/

    }
}