#include <iostream>
using namespace std;

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

char map[32][32];
int w, h;

int counter;

void count(int x, int y) {
	map[x][y] = '#';
	counter++;
	for (int i = 0; i < 4; i++) {
		int nextX = x + dx[i];
		int nextY = y + dy[i];
		if ((0 <= nextX && nextX < w) && (0 <= nextY && nextY < h)) {
			if (map[nextX][nextY] != '#') {
				count(nextX, nextY);
			}
		}
	}
}

int main() {
	int startX, startY;
	while (cin >> w >> h, w || h) {
		counter = 0;
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				cin >> map[j][i];
				if (map[j][i] == '@') {
					startX = j;
					startY = i;
				}
			}
		}
		count(startX, startY);
		cout << counter << endl;
	}
}
### Prompt

Your challenge is to write 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;

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

char map[32][32];
int w, h;

int counter;

void count(int x, int y) {
	map[x][y] = '#';
	counter++;
	for (int i = 0; i < 4; i++) {
		int nextX = x + dx[i];
		int nextY = y + dy[i];
		if ((0 <= nextX && nextX < w) && (0 <= nextY && nextY < h)) {
			if (map[nextX][nextY] != '#') {
				count(nextX, nextY);
			}
		}
	}
}

int main() {
	int startX, startY;
	while (cin >> w >> h, w || h) {
		counter = 0;
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				cin >> map[j][i];
				if (map[j][i] == '@') {
					startX = j;
					startY = i;
				}
			}
		}
		count(startX, startY);
		cout << counter << endl;
	}
}
```