#include <bits/stdc++.h>
using namespace	std;
static int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
int h,w,r;
string s[20];
void dfs(int y,int x){
  r++;
  s[y][x]='#';
  for(int i=0;i<4;i++)
      if(0<=y+dy[i]&&y+dy[i]<h&&0<=x+dx[i]&&x+dx[i]<w)
	if(s[y+dy[i]][x+dx[i]]=='.') dfs(y+dy[i],x+dx[i]);
}
int main(){

  int y,x;
  for(;cin>>w>>h,w;){
  r=0;
  for(int i=0;i<h;i++) cin>>s[i];
  for(int i=0;i<h;i++) for(int j=0;j<w;j++) if(s[i][j]=='@') y=i,x=j;
  dfs(y,x);
  cout<<r<<endl;
  }
  return 0;
}
### 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 <bits/stdc++.h>
using namespace	std;
static int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
int h,w,r;
string s[20];
void dfs(int y,int x){
  r++;
  s[y][x]='#';
  for(int i=0;i<4;i++)
      if(0<=y+dy[i]&&y+dy[i]<h&&0<=x+dx[i]&&x+dx[i]<w)
	if(s[y+dy[i]][x+dx[i]]=='.') dfs(y+dy[i],x+dx[i]);
}
int main(){

  int y,x;
  for(;cin>>w>>h,w;){
  r=0;
  for(int i=0;i<h;i++) cin>>s[i];
  for(int i=0;i<h;i++) for(int j=0;j<w;j++) if(s[i][j]=='@') y=i,x=j;
  dfs(y,x);
  cout<<r<<endl;
  }
  return 0;
}
```