#include <bits/stdc++.h>

using namespace std;

int W, H;
char room[20][20];

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

int sx, sy;

int dfs(int x, int y)
{
  room[y][x] = '#';
  int cnt = 1;
  for(int i = 0; i < 4; i++){
    int nx = x + dx[i], ny = y + dy[i];
    if(0 <= nx && nx < W && 0 <= ny && ny < H && room[ny][nx] != '#'){
      cnt += dfs(nx, ny);
    }
  }
  return cnt;
}

int main()
{
  while(cin >> W >> H, W || H){
    for(int i = 0; i < H; i++){
      for(int j = 0; j < W; j++){
	cin >> room[i][j];
	if(room[i][j] == '@') sx = j, sy = i;
      }
    }
    cout << dfs(sx, sy) << endl;
  }
  return 0;
}
### 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 <bits/stdc++.h>

using namespace std;

int W, H;
char room[20][20];

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

int sx, sy;

int dfs(int x, int y)
{
  room[y][x] = '#';
  int cnt = 1;
  for(int i = 0; i < 4; i++){
    int nx = x + dx[i], ny = y + dy[i];
    if(0 <= nx && nx < W && 0 <= ny && ny < H && room[ny][nx] != '#'){
      cnt += dfs(nx, ny);
    }
  }
  return cnt;
}

int main()
{
  while(cin >> W >> H, W || H){
    for(int i = 0; i < H; i++){
      for(int j = 0; j < W; j++){
	cin >> room[i][j];
	if(room[i][j] == '@') sx = j, sy = i;
      }
    }
    cout << dfs(sx, sy) << endl;
  }
  return 0;
}
```