#include <iostream>
using namespace std;
int w,h;
int tx[4] = {0,-1,0,1};
int ty[4] = {-1,0,1,0};
char map[21][21];

int slv(int x,int y){
    map[y][x]='#';
    int ans=0;
    for(int i=0;i<4;++i)
        if( x+tx[i]!=-1 && y+ty[i]!=-1 && x+tx[i]!=w && y+ty[i]!=h)
            if( map[y+ty[i]][x+tx[i]] == '.' )
                ans+=slv( x+tx[i], y+ty[i] ) + 1 ;
    return ans;
}

int main()
{
    int ix,iy;
    while(1){
        cin >> w >> h;
        if(w==0&&h==0)break;
        for(int i=0;i<h;++i)
            for(int j=0;j<w;++j){
                cin >> map[i][j];
                if(map[i][j]=='@'){ix=j;iy=i;}
            }
        cout << slv( ix , iy ) + 1 << endl;
    }
    return 0;
}
### 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 <iostream>
using namespace std;
int w,h;
int tx[4] = {0,-1,0,1};
int ty[4] = {-1,0,1,0};
char map[21][21];

int slv(int x,int y){
    map[y][x]='#';
    int ans=0;
    for(int i=0;i<4;++i)
        if( x+tx[i]!=-1 && y+ty[i]!=-1 && x+tx[i]!=w && y+ty[i]!=h)
            if( map[y+ty[i]][x+tx[i]] == '.' )
                ans+=slv( x+tx[i], y+ty[i] ) + 1 ;
    return ans;
}

int main()
{
    int ix,iy;
    while(1){
        cin >> w >> h;
        if(w==0&&h==0)break;
        for(int i=0;i<h;++i)
            for(int j=0;j<w;++j){
                cin >> map[i][j];
                if(map[i][j]=='@'){ix=j;iy=i;}
            }
        cout << slv( ix , iy ) + 1 << endl;
    }
    return 0;
}
```