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
#include<iostream>
#include<vector> 
using namespace std;

int x,y;
char ti[30][30];
const int dx[]={1,0,-1,0}, dy[]={0,-1,0,1};
int sum=0;

bool valid(int i,int j)
{
	return(i>=0&&i<y&&j>=0&&j<x);
}

void dfs(int m,int n)
{
	sum++;
	ti[m][n]='#';
	for(int i=0;i<4;i++)
	{
		int x,y;
		x=m+dx[i];
		y=n+dy[i];
		if(valid(x,y) && ti[x][y]=='.')
		  dfs(x,y);
	}
}

int main()
{
	int m,n;
	while(cin >> x >> y)
	{
		sum=0;
		if(!x&&!y)
		  break;
		for(int i=0;i<y;i++)
		   for(int j=0;j<x;j++)
		   {
		   	    cin >> ti[i][j];
		   	    if(ti[i][j]=='@')
		   	    {
		   	    	m=i;
		   	    	n=j;
				}
		   }
		dfs(m,n);
	    cout << sum << endl;
	}
}
