#include <iostream>
using namespace std;

#define BLACK '.'
#define RED '#'
#define ST '@'

int w,h, ans=0;
char t[100][100];

int rec(int x, int y){
  if( x<0 || x>=h ) return 0;
  if( y<0 || y>=w ) return 0;

  if( t[x][y] == RED ) return 0;
  t[x][y] = RED;

  int dx[4] = {0,0,1,-1};
  int dy[4] = {1,-1,0,0};
  int r=1;
  for( int d=0;d<4;d++ )
     r+= rec(x+dx[d], y+dy[d]);
  return r;
}
int main(){
  while( cin>>w >> h, w){
    int si, sj;
    for( int i=0;i<h;i++ )
      for( int j=0;j<w;j++) {
	cin >> t[i][j];
	if( t[i][j] == ST ){
	  si=i; sj=j;
	}
      }
    cout << rec(si,sj) << endl;
  }
}
### Prompt

Your challenge is to write 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>
using namespace std;

#define BLACK '.'
#define RED '#'
#define ST '@'

int w,h, ans=0;
char t[100][100];

int rec(int x, int y){
  if( x<0 || x>=h ) return 0;
  if( y<0 || y>=w ) return 0;

  if( t[x][y] == RED ) return 0;
  t[x][y] = RED;

  int dx[4] = {0,0,1,-1};
  int dy[4] = {1,-1,0,0};
  int r=1;
  for( int d=0;d<4;d++ )
     r+= rec(x+dx[d], y+dy[d]);
  return r;
}
int main(){
  while( cin>>w >> h, w){
    int si, sj;
    for( int i=0;i<h;i++ )
      for( int j=0;j<w;j++) {
	cin >> t[i][j];
	if( t[i][j] == ST ){
	  si=i; sj=j;
	}
      }
    cout << rec(si,sj) << endl;
  }
}
```