#include<bits/stdc++.h>
using namespace std;
char m[21][21];
int cnt=1,w,h;
void dfs(int x,int y){
  int i,j;
  m[y][x]='#';
  for(i=-1;i<=1;++i){
    if(y+i<0) continue;
    if(m[y+i][x]=='.'){
      cnt++;
      dfs(x,y+i);
    }
  }
  for(j=-1;j<=1;++j){
    if(x+j<0) continue;
    if(m[y][x+j]=='.'){
      cnt++;
      dfs(x+j,y);
    }
  }
}
int main(){
  int i,j,x,y;
  while(cin>>w>>h){
    if(w==0&&h==0) break;
    for(i=0;i<21;++i)
      for(j=0;j<21;++j)
	m[i][j]=0;
    for(i=0;i<h;++i){
      for(j=0;j<w;++j){
	cin>>m[i][j];
	if(m[i][j]=='@'){
	  x=j;
	  y=i;
	}
      }
    }
    cnt=1;
    dfs(x,y);
    cout<<cnt<<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<bits/stdc++.h>
using namespace std;
char m[21][21];
int cnt=1,w,h;
void dfs(int x,int y){
  int i,j;
  m[y][x]='#';
  for(i=-1;i<=1;++i){
    if(y+i<0) continue;
    if(m[y+i][x]=='.'){
      cnt++;
      dfs(x,y+i);
    }
  }
  for(j=-1;j<=1;++j){
    if(x+j<0) continue;
    if(m[y][x+j]=='.'){
      cnt++;
      dfs(x+j,y);
    }
  }
}
int main(){
  int i,j,x,y;
  while(cin>>w>>h){
    if(w==0&&h==0) break;
    for(i=0;i<21;++i)
      for(j=0;j<21;++j)
	m[i][j]=0;
    for(i=0;i<h;++i){
      for(j=0;j<w;++j){
	cin>>m[i][j];
	if(m[i][j]=='@'){
	  x=j;
	  y=i;
	}
      }
    }
    cnt=1;
    dfs(x,y);
    cout<<cnt<<endl;
  }
  return 0;
}
```