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

int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};

char f[50][50];
bool v[50][50];

int M,N;

int dfs(int x, int y)
{
	int ret=1;
	v[x][y]=1;

	for(int i=0; i<4; i++)
	{
		int tx=x+dx[i], ty=y+dy[i];
		if(tx<0||tx>=M||ty<0||ty>=N) continue;
		if(f[tx][ty]=='#') continue;
		if(v[tx][ty]) continue;

		ret+=dfs(tx,ty);
	}

	return ret;
}

int main()
{
	while(cin >> M >> N, (M||N))
	{
		int sx,sy;
		memset(v,0,sizeof(v));
		for(int i=0; i<N; i++)
		for(int j=0; j<M; j++)
		{
			cin >> f[j][i];
			if(f[j][i]=='@')
			{
				sx=j;
				sy=i;
			}
		}

		cout << dfs(sx,sy) << endl;
	}
}
### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <iostream>
#include <cstring>
using namespace std;

int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};

char f[50][50];
bool v[50][50];

int M,N;

int dfs(int x, int y)
{
	int ret=1;
	v[x][y]=1;

	for(int i=0; i<4; i++)
	{
		int tx=x+dx[i], ty=y+dy[i];
		if(tx<0||tx>=M||ty<0||ty>=N) continue;
		if(f[tx][ty]=='#') continue;
		if(v[tx][ty]) continue;

		ret+=dfs(tx,ty);
	}

	return ret;
}

int main()
{
	while(cin >> M >> N, (M||N))
	{
		int sx,sy;
		memset(v,0,sizeof(v));
		for(int i=0; i<N; i++)
		for(int j=0; j<M; j++)
		{
			cin >> f[j][i];
			if(f[j][i]=='@')
			{
				sx=j;
				sy=i;
			}
		}

		cout << dfs(sx,sy) << endl;
	}
}
```