#include <cstdio>

using namespace std;

#define rep(i,x) for(int i=0;i<(x);++i)

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

char mp[20][22];
int H, W;

void dfs(int y, int x)
{
	if (y < 0 || y >= H || x < 0 || x >= W || mp[y][x] == '#' || mp[y][x] == ' ') return;

	mp[y][x] = ' ';

	rep(i, 4) dfs(y + dy[i], x + dx[i]);
}

int main()
{
	while (scanf("%d %d", &W, &H), H || W) {
		int sy, sx;

		rep(i, H) rep(j, W) {
			scanf("\n%c", &mp[i][j]);
			if (mp[i][j] == '@') sy = i, sx = j, mp[i][j] = '.';
		}

		dfs(sy, sx);

		int cnt = 0;

		rep(i, H) rep(j, W) if (mp[i][j] == ' ') cnt++;

		printf("%d\n", cnt);
	}
}
### Prompt

Please formulate a cpp solution to 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 <cstdio>

using namespace std;

#define rep(i,x) for(int i=0;i<(x);++i)

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

char mp[20][22];
int H, W;

void dfs(int y, int x)
{
	if (y < 0 || y >= H || x < 0 || x >= W || mp[y][x] == '#' || mp[y][x] == ' ') return;

	mp[y][x] = ' ';

	rep(i, 4) dfs(y + dy[i], x + dx[i]);
}

int main()
{
	while (scanf("%d %d", &W, &H), H || W) {
		int sy, sx;

		rep(i, H) rep(j, W) {
			scanf("\n%c", &mp[i][j]);
			if (mp[i][j] == '@') sy = i, sx = j, mp[i][j] = '.';
		}

		dfs(sy, sx);

		int cnt = 0;

		rep(i, H) rep(j, W) if (mp[i][j] == ' ') cnt++;

		printf("%d\n", cnt);
	}
}
```