
#include <iostream>

using namespace std;

int w, h;
string f[20];
int cnt;

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

void rec(int x, int y) {
  f[y][x] = '#';
  cnt++;

  for (int i=0; i<4; i++) {
    int mx = x + dx[i];
    int my = y + dy[i];

    if (mx < 0 || w <= mx) { continue; }
    if (my < 0 || h <= my) { continue; }
    if (f[my][mx] != '.') { continue; }

    rec(mx, my);
  }
}

int main() {
  while (cin >> w >> h, (w||h)) {
    for (int i=0; i<h; i++) {
      cin >> f[i];
    }

    cnt = 0;
    for (int y=0; y<h; y++) {
      for (int x=0; x<w; x++) {
        if (f[y][x] == '@') { rec(x, y); }
      }
    }

    cout << cnt << endl;
  }

  return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 w, h;
string f[20];
int cnt;

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

void rec(int x, int y) {
  f[y][x] = '#';
  cnt++;

  for (int i=0; i<4; i++) {
    int mx = x + dx[i];
    int my = y + dy[i];

    if (mx < 0 || w <= mx) { continue; }
    if (my < 0 || h <= my) { continue; }
    if (f[my][mx] != '.') { continue; }

    rec(mx, my);
  }
}

int main() {
  while (cin >> w >> h, (w||h)) {
    for (int i=0; i<h; i++) {
      cin >> f[i];
    }

    cnt = 0;
    for (int y=0; y<h; y++) {
      for (int x=0; x<w; x++) {
        if (f[y][x] == '@') { rec(x, y); }
      }
    }

    cout << cnt << endl;
  }

  return 0;
}
```