#include<iostream>
using namespace std;
#define r(i,n) for(int i=0;i<n;i++)
int ans=1,n,m;
char map[30][30];

void d(int x,int y){
  if(x==m||y==n||x<0||y<0)return;
  if(map[x][y]=='.')ans++;
  if(map[x][y]=='#')return;
  map[x][y]='#';
  d(x+1,y),d(x-1,y),d(x,y+1),d(x,y-1);
}

int main(){
  while(cin>>n>>m,n){
  int x,y;
  ans=1;
  r(i,m)r(j,n){
    cin>>map[i][j];
    if(map[i][j]=='@')x=i,y=j;
  }
  d(x,y);
  cout<<ans<<endl;
  }
  return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define r(i,n) for(int i=0;i<n;i++)
int ans=1,n,m;
char map[30][30];

void d(int x,int y){
  if(x==m||y==n||x<0||y<0)return;
  if(map[x][y]=='.')ans++;
  if(map[x][y]=='#')return;
  map[x][y]='#';
  d(x+1,y),d(x-1,y),d(x,y+1),d(x,y-1);
}

int main(){
  while(cin>>n>>m,n){
  int x,y;
  ans=1;
  r(i,m)r(j,n){
    cin>>map[i][j];
    if(map[i][j]=='@')x=i,y=j;
  }
  d(x,y);
  cout<<ans<<endl;
  }
  return 0;
}
```