#include <iostream>
#include <string>
using namespace	std;
int result,w,h;
int d1[4]={-1,0,1,0},d2[4]={0,1,0,-1};
string a[21];

void dfs(int x,int y){

  int i;

  result++;
  a[y][x]='#';
  for(i=0;i<4;i++) if(0<=y+d2[i] && y+d2[i]<h && 0<=x+d1[i] && x+d1[i]<w) if(a[y+d2[i]][x+d1[i]] == '.') dfs(x+d1[i],y+d2[i]);

}

int main(){

  int x1,y1,i,j;

  while(cin>>w>>h,w){

    result=0;
    for(i=0;i<h;i++) cin>>a[i];

    for(i=0;i<h;i++) for(j=0;j<w;j++) if(a[i][j] == '@'){
          x1=j;
          y1=i;
	}

    dfs(x1,y1);

    cout<<result<<endl;


  }

  return 0;

}
### 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 <iostream>
#include <string>
using namespace	std;
int result,w,h;
int d1[4]={-1,0,1,0},d2[4]={0,1,0,-1};
string a[21];

void dfs(int x,int y){

  int i;

  result++;
  a[y][x]='#';
  for(i=0;i<4;i++) if(0<=y+d2[i] && y+d2[i]<h && 0<=x+d1[i] && x+d1[i]<w) if(a[y+d2[i]][x+d1[i]] == '.') dfs(x+d1[i],y+d2[i]);

}

int main(){

  int x1,y1,i,j;

  while(cin>>w>>h,w){

    result=0;
    for(i=0;i<h;i++) cin>>a[i];

    for(i=0;i<h;i++) for(j=0;j<w;j++) if(a[i][j] == '@'){
          x1=j;
          y1=i;
	}

    dfs(x1,y1);

    cout<<result<<endl;


  }

  return 0;

}
```