#include <iostream>
using namespace std;

int masu[20][20];
int W,H;

int dfs(int x,int y){
     if(x>=W||y>=H||x<0||y<0)return 0;
     if(masu[x][y]==1)return 0;
     masu[x][y]=1;
     return dfs(x+1,y)+dfs(x,y+1)+dfs(x-1,y)+dfs(x,y-1)+1;
}

int main(){
     while(1){
	  cin>>W>>H;
	  if(W==0)break;
	  int pos[2];
	  for(int j=0;j<H;j++){
	       string temp;
	       cin>>temp;
	       for(int i=0;i<W;i++){
		    switch(temp[i]){
		    case '.':
			 masu[i][j]=0;
			 break;
		    case '@':
			 masu[i][j]=0;
			 pos[0]=i;pos[1]=j;
			 break;
		    case '#':
			 masu[i][j]=1;
			 break;
		    }
	       }
	  }
	  int ans=dfs(pos[0],pos[1]);
	  cout<<ans<<endl;
     }
}
### Prompt

Please create a solution in Cpp 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 <iostream>
using namespace std;

int masu[20][20];
int W,H;

int dfs(int x,int y){
     if(x>=W||y>=H||x<0||y<0)return 0;
     if(masu[x][y]==1)return 0;
     masu[x][y]=1;
     return dfs(x+1,y)+dfs(x,y+1)+dfs(x-1,y)+dfs(x,y-1)+1;
}

int main(){
     while(1){
	  cin>>W>>H;
	  if(W==0)break;
	  int pos[2];
	  for(int j=0;j<H;j++){
	       string temp;
	       cin>>temp;
	       for(int i=0;i<W;i++){
		    switch(temp[i]){
		    case '.':
			 masu[i][j]=0;
			 break;
		    case '@':
			 masu[i][j]=0;
			 pos[0]=i;pos[1]=j;
			 break;
		    case '#':
			 masu[i][j]=1;
			 break;
		    }
	       }
	  }
	  int ans=dfs(pos[0],pos[1]);
	  cout<<ans<<endl;
     }
}
```