#include<stdio.h>
char a[21][21];
int c;
void dfs(int y,int x){
  c++;
  a[y][x]=0;
  if(y!=0)
    if(a[y-1][x]=='.')
      dfs(y-1,x);
  if(y!=20)
    if(a[y+1][x]=='.')
      dfs(y+1,x);
  if(x!=0)
    if(a[y][x-1]=='.')
      dfs(y,x-1);
  if(x!=20)
    if(a[y][x+1]=='.')
      dfs(y,x+1);
}
int main(){
  int w,h,i,j;
  char b[21];
  while(1){
    scanf("%d%d",&w,&h);
    if(w==0&&h==0)break;
    for(i=0;i<21;i++){
      for(j=0;j<21;j++){
	a[i][j]=0;
      }
    }
    for(i=0;i<h;i++){
	scanf("%s",b);
	for(j=0;b[j]!='\0';j++){
	  a[i][j]=b[j];
	}
    }
    c=0;
    for(i=0;i<h;i++){
      for(j=0;j<w;j++){
	if(a[i][j]=='@')dfs(i,j);
      }
    }
    printf("%d\n",c);

   
  }
  return 0;
}
### Prompt

In CPP, your task is to solve 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<stdio.h>
char a[21][21];
int c;
void dfs(int y,int x){
  c++;
  a[y][x]=0;
  if(y!=0)
    if(a[y-1][x]=='.')
      dfs(y-1,x);
  if(y!=20)
    if(a[y+1][x]=='.')
      dfs(y+1,x);
  if(x!=0)
    if(a[y][x-1]=='.')
      dfs(y,x-1);
  if(x!=20)
    if(a[y][x+1]=='.')
      dfs(y,x+1);
}
int main(){
  int w,h,i,j;
  char b[21];
  while(1){
    scanf("%d%d",&w,&h);
    if(w==0&&h==0)break;
    for(i=0;i<21;i++){
      for(j=0;j<21;j++){
	a[i][j]=0;
      }
    }
    for(i=0;i<h;i++){
	scanf("%s",b);
	for(j=0;b[j]!='\0';j++){
	  a[i][j]=b[j];
	}
    }
    c=0;
    for(i=0;i<h;i++){
      for(j=0;j<w;j++){
	if(a[i][j]=='@')dfs(i,j);
      }
    }
    printf("%d\n",c);

   
  }
  return 0;
}
```