#include<iostream>

using namespace std;

void dfs(int mx,int my);

	char math[21][21];
	const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
	int cnt;
	int x,y;
	int sx,sy;
	
int main(){
	
	while(cin>>x>>y,x&&y){
	    cnt=0;
		for(int i=0;i<y;i++){
			for(int j=0;j<x;j++){
				cin>>math[i][j];
				if(math[i][j]=='@'){
					sx=j;
					sy=i;
				}
			}
		}
		dfs(sx,sy);
		cout<<cnt+1<<endl;
	}
}

void dfs(int mx,int my){
	math[my][mx]='T';
	//移動する方向
	for(int i=0;i<4;i++){
		int nx=mx+dx[i],ny=my+dy[i];
		
		if(0<=nx && nx<x && 0<=ny && ny<y &&
		   (math[ny][nx]=='.' ||math[ny][nx]=='@')
		  )
		  {
			cnt++;
			dfs(nx,ny);
		}
	}
	return ;
}
### 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;

void dfs(int mx,int my);

	char math[21][21];
	const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
	int cnt;
	int x,y;
	int sx,sy;
	
int main(){
	
	while(cin>>x>>y,x&&y){
	    cnt=0;
		for(int i=0;i<y;i++){
			for(int j=0;j<x;j++){
				cin>>math[i][j];
				if(math[i][j]=='@'){
					sx=j;
					sy=i;
				}
			}
		}
		dfs(sx,sy);
		cout<<cnt+1<<endl;
	}
}

void dfs(int mx,int my){
	math[my][mx]='T';
	//移動する方向
	for(int i=0;i<4;i++){
		int nx=mx+dx[i],ny=my+dy[i];
		
		if(0<=nx && nx<x && 0<=ny && ny<y &&
		   (math[ny][nx]=='.' ||math[ny][nx]=='@')
		  )
		  {
			cnt++;
			dfs(nx,ny);
		}
	}
	return ;
}
```