#include <iostream>
using namespace std;
char f[22][22];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0}; 
int w,h,sx,sy;
void dfs(int x,int y)
{
	if(f[x][y]!='.')return;
	f[x][y]='7';
	for(int i=0;i<4;i++)
	{
		int nx = x + dx[i];
		int ny = y + dy[i];
		if(nx<0||nx>=h||ny<0||ny>=w)continue;
		if(f[nx][ny]!='.')continue;
		dfs(nx,ny);
	}
	return;
}
int main()
{
	while(1)
	{
		cin >> w >> h;
		if(w==0&&h==0)break;
		for(int i=0;i<h;i++)
		{
			for(int j=0;j<w;j++)
			{
				cin >> f[i][j];
				if(f[i][j]=='@')
				{
					sx = i;
					sy = j;
					f[i][j]='.';
				}
			}
		}
		dfs(sx,sy);
		int ans = 0;
		for(int i=0;i<h;i++)
		{
			for(int j=0;j<w;j++)
			{
				if(f[i][j]=='7')ans++;
			}
		}
		cout << ans << endl;
	}
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <iostream>
using namespace std;
char f[22][22];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0}; 
int w,h,sx,sy;
void dfs(int x,int y)
{
	if(f[x][y]!='.')return;
	f[x][y]='7';
	for(int i=0;i<4;i++)
	{
		int nx = x + dx[i];
		int ny = y + dy[i];
		if(nx<0||nx>=h||ny<0||ny>=w)continue;
		if(f[nx][ny]!='.')continue;
		dfs(nx,ny);
	}
	return;
}
int main()
{
	while(1)
	{
		cin >> w >> h;
		if(w==0&&h==0)break;
		for(int i=0;i<h;i++)
		{
			for(int j=0;j<w;j++)
			{
				cin >> f[i][j];
				if(f[i][j]=='@')
				{
					sx = i;
					sy = j;
					f[i][j]='.';
				}
			}
		}
		dfs(sx,sy);
		int ans = 0;
		for(int i=0;i<h;i++)
		{
			for(int j=0;j<w;j++)
			{
				if(f[i][j]=='7')ans++;
			}
		}
		cout << ans << endl;
	}
	return 0;
}
```