#include <cstdio>


char board[20][20] = {};
int W, H;

int count(int x, int y) {
	if (board[x][y] == '#') return 0;
	if (x < 0 || W <= x) return 0;
	if (y < 0 || H <= y) return 0;

	board[x][y] = '#';

	return 1 +
		count(x - 1, y) +
		count(x + 1, y) +
		count(x, y - 1) +
		count(x, y + 1);
}

int main() {
	while (true) {

		scanf("%d %d", &W, &H);

		if (W == 0) break;

		int cx, cy;

		for (int y = 0; y < H; y++) {
			char buf[21];
			scanf("%s", buf);
			for (int x = 0; x < W; x++) {
				board[x][y] = buf[x];
				if (buf[x] == '@') {
					cx = x; cy = y;
				}
			}
		}

		printf("%d\n", count(cx, cy));
	}

	return 0;
}
### 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 <cstdio>


char board[20][20] = {};
int W, H;

int count(int x, int y) {
	if (board[x][y] == '#') return 0;
	if (x < 0 || W <= x) return 0;
	if (y < 0 || H <= y) return 0;

	board[x][y] = '#';

	return 1 +
		count(x - 1, y) +
		count(x + 1, y) +
		count(x, y - 1) +
		count(x, y + 1);
}

int main() {
	while (true) {

		scanf("%d %d", &W, &H);

		if (W == 0) break;

		int cx, cy;

		for (int y = 0; y < H; y++) {
			char buf[21];
			scanf("%s", buf);
			for (int x = 0; x < W; x++) {
				board[x][y] = buf[x];
				if (buf[x] == '@') {
					cx = x; cy = y;
				}
			}
		}

		printf("%d\n", count(cx, cy));
	}

	return 0;
}
```