#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int cnt;
int w, h;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
char tile[200][200];

void dfs(int x, int y);

int main()
{
	int x, y;

	while (true){
		scanf("%d %d", &w, &h);

		if (w == 0){
			break;
		}

		for (int i = 0; i < h; i++){
			scanf("%s", tile[i]);

			for (int j = 0; j < w; j++){
				if (tile[i][j] == '@'){
					y = i;
					x = j;
				}
			}
		}

		cnt = 0;
		dfs(x, y);

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

	return (0);
}

void dfs(int x, int y)
{
	tile[y][x] = '#';
	cnt++;

	for (int i = 0; i < 4; i++){
		int nx = x + dx[i];
		int ny = y + dy[i];

		if (nx >= 0 && nx < w && ny >= 0 && ny < h && tile[ny][nx] == '.'){
			dfs(nx, ny);
		}
	}
}
### Prompt

Please provide a cpp coded solution 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>
#include <cstdio>
#include <algorithm>

using namespace std;

int cnt;
int w, h;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
char tile[200][200];

void dfs(int x, int y);

int main()
{
	int x, y;

	while (true){
		scanf("%d %d", &w, &h);

		if (w == 0){
			break;
		}

		for (int i = 0; i < h; i++){
			scanf("%s", tile[i]);

			for (int j = 0; j < w; j++){
				if (tile[i][j] == '@'){
					y = i;
					x = j;
				}
			}
		}

		cnt = 0;
		dfs(x, y);

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

	return (0);
}

void dfs(int x, int y)
{
	tile[y][x] = '#';
	cnt++;

	for (int i = 0; i < 4; i++){
		int nx = x + dx[i];
		int ny = y + dy[i];

		if (nx >= 0 && nx < w && ny >= 0 && ny < h && tile[ny][nx] == '.'){
			dfs(nx, ny);
		}
	}
}
```