#include <iostream>
#define MAX 20
using namespace std;

typedef pair<int,int> P;
int X, Y, ans;

char field[MAX][MAX];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};

void bfs(int px, int py)
{
     field[px][py] = '#';
     ans++;
     for (int i = 0; i < 4; i++) {
	  int x = px + dx[i], y = py + dy[i];
	  if (0 <= x && x < X && 0 <= y && y < Y && field[x][y] == '.') bfs(x,y);
     }
}

int main(void)
{
     int x, y;
     while (cin >> Y >> X && X && Y) {
	  for (int i = 0; i < X; i++)
	       for (int j = 0; j < Y; j++) {
		    cin >> field[i][j];
		    if (field[i][j] == '@') {
			 x = i;
			 y = j;
		    }
	       }
	  ans = 0;
	  bfs(x,y);
	  cout << ans << endl;
     }

     return 0;
}
### Prompt

In cpp, your task is to solve 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>
#define MAX 20
using namespace std;

typedef pair<int,int> P;
int X, Y, ans;

char field[MAX][MAX];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};

void bfs(int px, int py)
{
     field[px][py] = '#';
     ans++;
     for (int i = 0; i < 4; i++) {
	  int x = px + dx[i], y = py + dy[i];
	  if (0 <= x && x < X && 0 <= y && y < Y && field[x][y] == '.') bfs(x,y);
     }
}

int main(void)
{
     int x, y;
     while (cin >> Y >> X && X && Y) {
	  for (int i = 0; i < X; i++)
	       for (int j = 0; j < Y; j++) {
		    cin >> field[i][j];
		    if (field[i][j] == '@') {
			 x = i;
			 y = j;
		    }
	       }
	  ans = 0;
	  bfs(x,y);
	  cout << ans << endl;
     }

     return 0;
}
```