#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

#define rep(i,a) for(int i = 0 ; i < a ; i ++)
#define loop(i,a,b) for(int i = a ; i < b ; i ++)

int ans;
char maze[27][27];
void dfs(int x,int y){
	int dx[4] = {1,-1, 0, 0};
	int dy[4] = {0, 0, 1, -1};
	maze[x][y]='#';
	rep(i,4){
		int nx = dx[i] + x;
		int ny = dy[i] + y;
		if(maze[nx][ny]=='.'){
			ans++;
			dfs(nx,ny);
		}
	}
}

int main(void){
	int w,h;
	while(cin>>w>>h){
		rep(i,25)rep(j,25)
			maze[i][j]='#';
		ans=0;
		if(w==0&&h==0)return 0;
		rep(i,h)rep(j,w)
			cin>>maze[i][j];
		rep(i,h)rep(j,w){
			if(maze[i][j]=='@'){
				dfs(i,j);
				cout<<ans+1<<endl;
			}
		}
	}
}
### Prompt

Your task is to create 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<iostream>
#include<vector>
#include<algorithm>

using namespace std;

#define rep(i,a) for(int i = 0 ; i < a ; i ++)
#define loop(i,a,b) for(int i = a ; i < b ; i ++)

int ans;
char maze[27][27];
void dfs(int x,int y){
	int dx[4] = {1,-1, 0, 0};
	int dy[4] = {0, 0, 1, -1};
	maze[x][y]='#';
	rep(i,4){
		int nx = dx[i] + x;
		int ny = dy[i] + y;
		if(maze[nx][ny]=='.'){
			ans++;
			dfs(nx,ny);
		}
	}
}

int main(void){
	int w,h;
	while(cin>>w>>h){
		rep(i,25)rep(j,25)
			maze[i][j]='#';
		ans=0;
		if(w==0&&h==0)return 0;
		rep(i,h)rep(j,w)
			cin>>maze[i][j];
		rep(i,h)rep(j,w){
			if(maze[i][j]=='@'){
				dfs(i,j);
				cout<<ans+1<<endl;
			}
		}
	}
}
```