#include <bits/stdc++.h>
using namespace std;
char maze[21][21];
#define r(z,x) for(int z=0;z<x;z++)
int ans,a,b;
void dfs(int x,int y){
	int dx[4]={0,-1,0,1},dy[4]={1,0,-1,0};
	maze[x][y]='#',ans++;
	r(i,4)if(maze[x+dx[i]][y+dy[i]]=='.')dfs(x+dx[i],y+dy[i]);
}
int main(){
	while(cin>>a>>b,a||b){
		ans=0;
		fill(maze[0],maze[20],'#');
		r(i,b)r(j,a)cin>>maze[i][j];
		r(i,b)r(j,a)if(maze[i][j]=='@')dfs(i,j),cout<<ans<<endl;
	}
}
### 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
#include <bits/stdc++.h>
using namespace std;
char maze[21][21];
#define r(z,x) for(int z=0;z<x;z++)
int ans,a,b;
void dfs(int x,int y){
	int dx[4]={0,-1,0,1},dy[4]={1,0,-1,0};
	maze[x][y]='#',ans++;
	r(i,4)if(maze[x+dx[i]][y+dy[i]]=='.')dfs(x+dx[i],y+dy[i]);
}
int main(){
	while(cin>>a>>b,a||b){
		ans=0;
		fill(maze[0],maze[20],'#');
		r(i,b)r(j,a)cin>>maze[i][j];
		r(i,b)r(j,a)if(maze[i][j]=='@')dfs(i,j),cout<<ans<<endl;
	}
}
```