Problem: There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.



Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

* '.' - a black tile
* '#' - a red tile
* '@' - a man on a black tile(appears exactly once in a data set)


The end of the input is indicated by a line consisting of two zeros.

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Examples

Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0


Output

45
59
6
13


Input

6 9
....#.
.....#
......
......
......
......
......
@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
.###
...@...
.###
..#.#..
..#.#..
0 0


Output

45
59
6
13

Solution (in JAVA):

Solution:
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

/**
 * @author yousack
 */
public class Main {
    static Scanner in = new Scanner(System.in);
        
    public static void main(String[] args) {
        while (true) {
            int W = in.nextInt();
            int H = in.nextInt();

            if (W == 0 && H == 0) {
                break;
            } else {
                char[][] field = new char[H][W];
                int r = 0, c = 0;

                for (int i = 0; i < H; i++) {
                    field[i] = in.next().toCharArray();
                    for (int j = 0; j < W; j++) {
                        if (field[i][j] == '@') {
                            field[i][j] = '.';
                            r = i;
                            c = j;
                        }
                    }
                }
                
                dfs(field, r, c);
            }
        }
    }
    
    public static void dfs(char[][] f, int r, int c) {
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.add(new int[]{r, c});
        int counter = 0;
        int[][] dirArray = new int[][]{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        
        while (!queue.isEmpty()) {
            int[] curPos = queue.poll();
            int curR = curPos[0], curC = curPos[1];
            if (f[curR][curC] == '.') {
                counter++;
                f[curR][curC] = '@';
                for (int[] next : dirArray) {
                    if (ok(curR + next[0], curC + next[1], f.length, f[0].length)) {
                        queue.add(new int[]{curR + next[0], curC + next[1]});
                    }
                }
            }
        }

        System.out.println(counter);
    }
    
    public static boolean ok(int posR, int posC, int boundR, int boundC) {
        return 0 <= posR && posR < boundR && 0 <= posC && posC < boundC;         
    }
}