#include <iostream>
using namespace std;

char field[30][30];
int res;
int w, h;
constexpr int dd[] = {0, 1, 0, -1, 0};

void dfs(int y, int x) {
  res++;
  field[y][x] = 'x';
  for (int i = 0; i < 4; ++i) {
    int ny = dd[i] + y, nx = dd[i + 1] + x;
    if (0 <= ny && ny < h && 0 <= nx && nx < w) {
      if (field[ny][nx] == '.') {
        dfs(ny, nx);
      }
    }
  }
}

int main() {
  while (cin >> w >> h, w || h) {
    int sy, sx;
    for (int i = 0; i < h; ++i) {
      for (int j = 0; j < w; ++j) {
        cin >> field[i][j];
        if (field[i][j] == '@') {
          sy = i;
          sx = j;
          field[i][j] = '.';
        }
      }
    }
    res = 0;
    dfs(sy, sx);
    cout << res << endl;
  }
}
### 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 <iostream>
using namespace std;

char field[30][30];
int res;
int w, h;
constexpr int dd[] = {0, 1, 0, -1, 0};

void dfs(int y, int x) {
  res++;
  field[y][x] = 'x';
  for (int i = 0; i < 4; ++i) {
    int ny = dd[i] + y, nx = dd[i + 1] + x;
    if (0 <= ny && ny < h && 0 <= nx && nx < w) {
      if (field[ny][nx] == '.') {
        dfs(ny, nx);
      }
    }
  }
}

int main() {
  while (cin >> w >> h, w || h) {
    int sy, sx;
    for (int i = 0; i < h; ++i) {
      for (int j = 0; j < w; ++j) {
        cin >> field[i][j];
        if (field[i][j] == '@') {
          sy = i;
          sx = j;
          field[i][j] = '.';
        }
      }
    }
    res = 0;
    dfs(sy, sx);
    cout << res << endl;
  }
}
```