#include<iostream>
using namespace std;
char map[114][514];
int h, w, ans;
void dfs(int i, int j) {
	ans++;
	map[i][j] = '#';
	if (map[i - 1][j] == '.')dfs(i - 1, j);
	if (map[i + 1][j] == '.')dfs(i + 1, j);
	if (map[i][j - 1] == '.')dfs(i, j - 1);
	if (map[i][j + 1] == '.')dfs(i, j + 1);
	return;
}
int main() {
	while (cin >> h >> w, h, w) {
		ans = 0;
		int x, y;
		for (int i = 0; i < 30; i++)for (int j = 0; j < 30; j++)map[i][j] = '#';
		for (int i = 1; i <= w; i++)for (int j = 1; j <= h; j++) {
			cin >> map[i][j];
			if (map[i][j] == '@') { x = i; y = j; }
		}
		dfs(x, y);
		cout << ans << endl;
	}
}
### 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<iostream>
using namespace std;
char map[114][514];
int h, w, ans;
void dfs(int i, int j) {
	ans++;
	map[i][j] = '#';
	if (map[i - 1][j] == '.')dfs(i - 1, j);
	if (map[i + 1][j] == '.')dfs(i + 1, j);
	if (map[i][j - 1] == '.')dfs(i, j - 1);
	if (map[i][j + 1] == '.')dfs(i, j + 1);
	return;
}
int main() {
	while (cin >> h >> w, h, w) {
		ans = 0;
		int x, y;
		for (int i = 0; i < 30; i++)for (int j = 0; j < 30; j++)map[i][j] = '#';
		for (int i = 1; i <= w; i++)for (int j = 1; j <= h; j++) {
			cin >> map[i][j];
			if (map[i][j] == '@') { x = i; y = j; }
		}
		dfs(x, y);
		cout << ans << endl;
	}
}
```