#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

void dfs(int, int);
int limx,limy,cnt=0,ni,nj;
char data[21][21];

int main(){
  int i,j;
  while(1){
    scanf("%d%d",&limx,&limy);
    if(limx==0 && limy==0)
      break;
    for(i=0;i<limy;i++)
      scanf("%s",data[i]);

    for(i=0;i<limy;i++)
      for(j=0;j<limx;j++)
	if(data[i][j]=='@') ni=i,nj=j;
    dfs(ni,nj);
    printf("%d\n",cnt);
    cnt=0;
  }
  return 0;
}

void dfs(int i, int j){
  int di[4]={0,-1,0,1};
  int dj[4]={1,0,-1,0};

  data[i][j]='#';
  cnt++;
  for(int r=0;r<4;r++){
    ni=i+di[r];
    nj=j+dj[r];
    if(data[ni][nj]=='.' && ni >= 0 && ni < limy && nj >= 0 && nj < limx) 
      dfs(ni,nj);
  }
}
### 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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

void dfs(int, int);
int limx,limy,cnt=0,ni,nj;
char data[21][21];

int main(){
  int i,j;
  while(1){
    scanf("%d%d",&limx,&limy);
    if(limx==0 && limy==0)
      break;
    for(i=0;i<limy;i++)
      scanf("%s",data[i]);

    for(i=0;i<limy;i++)
      for(j=0;j<limx;j++)
	if(data[i][j]=='@') ni=i,nj=j;
    dfs(ni,nj);
    printf("%d\n",cnt);
    cnt=0;
  }
  return 0;
}

void dfs(int i, int j){
  int di[4]={0,-1,0,1};
  int dj[4]={1,0,-1,0};

  data[i][j]='#';
  cnt++;
  for(int r=0;r<4;r++){
    ni=i+di[r];
    nj=j+dj[r];
    if(data[ni][nj]=='.' && ni >= 0 && ni < limy && nj >= 0 && nj < limx) 
      dfs(ni,nj);
  }
}
```