#include <stdio.h>
char c[1000][1000];
bool went[1000][1000];
int ans,n,m;
void go(int x,int y){
    if(x>=n||y>=m||x<0||y<0)return ;
    if(c[x][y]=='#')return;
    if(went[x][y])return;
    went[x][y]=true;
    go(x+1,y);
    go(x,y+1);
    go(x-1,y);
    go(x,y-1);
    ans++;
    return ;
}
int main(){
    bool f;
    while(1){
        scanf("%d%d",&m,&n);
        if(n==0&&m==0)return 0;
        f=true;
        ans=0;
        for(int i=0;i<n;i++)scanf("%s",c[i]);
        for(int i=0;i<n;i++)for(int j=0;j<m;j++)went[i][j]=false;
        for(int i=0;i<n&&f;i++)for(int j=0;j<m&&f;j++)if(c[i][j]=='@'){
            go(i,j);
            f=false;
        }
        printf("%d\n",ans);
    }
}
### 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 <stdio.h>
char c[1000][1000];
bool went[1000][1000];
int ans,n,m;
void go(int x,int y){
    if(x>=n||y>=m||x<0||y<0)return ;
    if(c[x][y]=='#')return;
    if(went[x][y])return;
    went[x][y]=true;
    go(x+1,y);
    go(x,y+1);
    go(x-1,y);
    go(x,y-1);
    ans++;
    return ;
}
int main(){
    bool f;
    while(1){
        scanf("%d%d",&m,&n);
        if(n==0&&m==0)return 0;
        f=true;
        ans=0;
        for(int i=0;i<n;i++)scanf("%s",c[i]);
        for(int i=0;i<n;i++)for(int j=0;j<m;j++)went[i][j]=false;
        for(int i=0;i<n&&f;i++)for(int j=0;j<m&&f;j++)if(c[i][j]=='@'){
            go(i,j);
            f=false;
        }
        printf("%d\n",ans);
    }
}
```