#include <cstdio>
#include <cstring>

char map[21][21];
int res;
int w,h;

void dfs(int sx, int sy) {
    if(sx<0 || sx>w-1 || sy<0 || sy>h-1) return;
    if(map[sy][sx] == '#') return;

    res++;
    map[sy][sx] = '#';

    dfs(sx+1,sy);
    dfs(sx,sy+1);
    dfs(sx-1,sy);
    dfs(sx,sy-1);
    return;
}

int main(void)
{
    int px,py;

    while(true)
    {
        res = 0;
        scanf("%d%d", &w, &h);
        if(!w && !h) break;
        for(int j=0; j < h; j++)
            scanf("%s", map[j]);

        for(int j=0; j < h; j++)
            for(int i=0; i < w; i++)
                if(map[j][i] == '@')
                    py = j, px =i;
        dfs(px, py);
        printf("%d\n", res);
    }

    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 <cstdio>
#include <cstring>

char map[21][21];
int res;
int w,h;

void dfs(int sx, int sy) {
    if(sx<0 || sx>w-1 || sy<0 || sy>h-1) return;
    if(map[sy][sx] == '#') return;

    res++;
    map[sy][sx] = '#';

    dfs(sx+1,sy);
    dfs(sx,sy+1);
    dfs(sx-1,sy);
    dfs(sx,sy-1);
    return;
}

int main(void)
{
    int px,py;

    while(true)
    {
        res = 0;
        scanf("%d%d", &w, &h);
        if(!w && !h) break;
        for(int j=0; j < h; j++)
            scanf("%s", map[j]);

        for(int j=0; j < h; j++)
            for(int i=0; i < w; i++)
                if(map[j][i] == '@')
                    py = j, px =i;
        dfs(px, py);
        printf("%d\n", res);
    }

    return 0;
}
```