7.31 D - Red and Black

                                    D - Red and Black

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).

Sample Input

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

Sample Output

45
59
6
13

#include<stdio.h>
#include<string.h>
char s[150][150];
int book[150][150];
int m,n,t;
int dfs(int x,int y)
{
	int i,j,k,tx,ty;
	int next[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
	
	if(x<0||y<0||x>=m||y>=n)
		return 0;	
		
	if(s[x][y]=='#')
	{
		return 0;
	}
	else
	{
		s[x][y]='#';
		return 1+dfs(x+1,y)+dfs(x-1,y)+dfs(x,y+1)+dfs(x,y-1);
	}

}
int main()
{
	int i,j,k,t,flag;
	while(scanf("%d%d",&n,&m),m+n!=0)
	{
		memset(book,0,sizeof(book));
		for(i=0;i<m;i++)
		{
			scanf("%s",s[i]);
		}
		for(i=0;i<m;i++)
		{
			for(j=0;j<n;j++)
			{
				if(s[i][j]=='@')
				{
					t=dfs(i,j);
				}
			}
		}
		printf("%d\n",t);
	}
	return 0;
}