#include <iostream>
using namespace std;
char m[21][21];
int w,h,ans;
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};

void solve(int x,int y) {
  m[y][x]='#';
  ans++;
  
  for(int i=0;i<4;i++) {
    int nx=x+dx[i],ny=y+dy[i];
    if(nx>=0 && nx<w && ny>=0 && ny<h && m[ny][nx]!='#') {
	solve(nx,ny);
    }
  }
}

int main() {
  int sx,sy;
  while(cin>>w>>h,w||h) {
    ans=0;
    for(int y=0;y<h;y++) {
      for(int x=0;x<w;x++) {
	cin>>m[y][x];
	if(m[y][x]=='@') {
	  sx=x;sy=y;
	}
      }
    }
    solve(sx,sy);
    cout<<ans<<endl;
  }

  return 0;
}
### Prompt

Create a solution in cpp for 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;
char m[21][21];
int w,h,ans;
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};

void solve(int x,int y) {
  m[y][x]='#';
  ans++;
  
  for(int i=0;i<4;i++) {
    int nx=x+dx[i],ny=y+dy[i];
    if(nx>=0 && nx<w && ny>=0 && ny<h && m[ny][nx]!='#') {
	solve(nx,ny);
    }
  }
}

int main() {
  int sx,sy;
  while(cin>>w>>h,w||h) {
    ans=0;
    for(int y=0;y<h;y++) {
      for(int x=0;x<w;x++) {
	cin>>m[y][x];
	if(m[y][x]=='@') {
	  sx=x;sy=y;
	}
      }
    }
    solve(sx,sy);
    cout<<ans<<endl;
  }

  return 0;
}
```