#include <iostream>
#include <string>

using namespace std;

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

string tile[20];

int w, h;

int dfs(int x, int y){
	int ret = 1;
	if (tile[x][y] == '#') return 0;
	tile[x][y] = '#';

	for (int i = 0; i < 4; i++){
		if (x + dx[i] < 0 || x + dx[i] >= h) continue;
		if (y + dy[i] < 0 || y + dy[i] >= w) continue;
		ret += dfs(x + dx[i], y + dy[i]);
	}
	return ret;
}

int main(){
	while (cin >> w >> h,w){
		int x, y;
		for (int i = 0; i < h; i++){
			cin >> tile[i];
			for (int j = 0; j < w; j++){
				if (tile[i][j] == '@'){
					x = i;
					y = j;
				}
			}
		}
		cout << dfs(x, y) << 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 <string>

using namespace std;

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

string tile[20];

int w, h;

int dfs(int x, int y){
	int ret = 1;
	if (tile[x][y] == '#') return 0;
	tile[x][y] = '#';

	for (int i = 0; i < 4; i++){
		if (x + dx[i] < 0 || x + dx[i] >= h) continue;
		if (y + dy[i] < 0 || y + dy[i] >= w) continue;
		ret += dfs(x + dx[i], y + dy[i]);
	}
	return ret;
}

int main(){
	while (cin >> w >> h,w){
		int x, y;
		for (int i = 0; i < h; i++){
			cin >> tile[i];
			for (int j = 0; j < w; j++){
				if (tile[i][j] == '@'){
					x = i;
					y = j;
				}
			}
		}
		cout << dfs(x, y) << endl;
	}
}
```