// PKU 1979
#include<cstdio>
#define rep( i, a ) for( int i = 0; i != (a); ++i )

int W, H;
char fld[20][20];
int ans;

const int dx[4] = { 0, 1, 0, -1 }, dy[4] = { -1, 0, 1, 0 };

void dfs( int x, int y )
{
	++ans;
	fld[y][x] = ' ';

	rep( i, 4 )
	{
		int nx = x + dx[i], ny = y + dy[i];
		
		if( nx >= 0 && nx < W && ny >= 0 && ny < H && fld[ny][nx] == '.' )
			dfs( nx, ny );
	}

	return;
}

int main()
{
	while( scanf( "%d%d", &W, &H ), W | H )
	{
		ans = 0;
		int sx, sy;

		rep( i, H )
		{
			getchar();

			rep( j, W )
			{
				fld[i][j] = getchar();
			
				if( fld[i][j] == '@' )
					sx = j, sy = i;
			}
		}

		dfs( sx, sy );

		printf( "%d\n", ans );
	}

	return 0;
}
### Prompt

Please formulate 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
// PKU 1979
#include<cstdio>
#define rep( i, a ) for( int i = 0; i != (a); ++i )

int W, H;
char fld[20][20];
int ans;

const int dx[4] = { 0, 1, 0, -1 }, dy[4] = { -1, 0, 1, 0 };

void dfs( int x, int y )
{
	++ans;
	fld[y][x] = ' ';

	rep( i, 4 )
	{
		int nx = x + dx[i], ny = y + dy[i];
		
		if( nx >= 0 && nx < W && ny >= 0 && ny < H && fld[ny][nx] == '.' )
			dfs( nx, ny );
	}

	return;
}

int main()
{
	while( scanf( "%d%d", &W, &H ), W | H )
	{
		ans = 0;
		int sx, sy;

		rep( i, H )
		{
			getchar();

			rep( j, W )
			{
				fld[i][j] = getchar();
			
				if( fld[i][j] == '@' )
					sx = j, sy = i;
			}
		}

		dfs( sx, sy );

		printf( "%d\n", ans );
	}

	return 0;
}
```