#include <iostream>
#include <string>

using namespace std;

/*??°?????????????????°*/
int W, H, ans;
int tile[20][20];

/*??¢??°??????????????????*/
void DPS(int x, int y);

int main() {
	int sx = 0, sy = 0;
	string str;

	while (cin >> W >> H, W, H) {
		/*??\???*/
		for (int i = 0; i < H; i++) {
			cin >> str;
			for (int j = 0; j < W; j++) {
				tile[i][j] = (str[j] == '#') ? 0 : 1;
				
				if (str[j] == '@') {
					sx = j;
					sy = i;
				}
			}
		}

		ans = 0;
		DPS(sx, sy);

		cout << ans << endl;
	}
}

void DPS(int x, int y) {
	if (tile[y][x]) {
		tile[y][x] = 0;
		ans++;
		
		if (x - 1 >= 0)	DPS(x - 1, y);
		if (x + 1 < W)	DPS(x + 1, y);
		if (y - 1 >= 0)	DPS(x, y - 1);
		if (y + 1 < H)	DPS(x, y + 1);
	}

	return;
}
### Prompt

Generate 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>
#include <string>

using namespace std;

/*??°?????????????????°*/
int W, H, ans;
int tile[20][20];

/*??¢??°??????????????????*/
void DPS(int x, int y);

int main() {
	int sx = 0, sy = 0;
	string str;

	while (cin >> W >> H, W, H) {
		/*??\???*/
		for (int i = 0; i < H; i++) {
			cin >> str;
			for (int j = 0; j < W; j++) {
				tile[i][j] = (str[j] == '#') ? 0 : 1;
				
				if (str[j] == '@') {
					sx = j;
					sy = i;
				}
			}
		}

		ans = 0;
		DPS(sx, sy);

		cout << ans << endl;
	}
}

void DPS(int x, int y) {
	if (tile[y][x]) {
		tile[y][x] = 0;
		ans++;
		
		if (x - 1 >= 0)	DPS(x - 1, y);
		if (x + 1 < W)	DPS(x + 1, y);
		if (y - 1 >= 0)	DPS(x, y - 1);
		if (y + 1 < H)	DPS(x, y + 1);
	}

	return;
}
```