#include <cstdio>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

int main(){
	for(int w,h;scanf("%d%d",&w,&h),w;){
		char B[20][21];
		rep(i,h) scanf("%s",B[i]);

		int ans=1;
		rep(_,h*w){
			rep(i,h) rep(j,w) if(B[i][j]=='@') {
				rep(k,4){
					int y=i+dy[k],x=j+dx[k];
					if(0<=y && y<h && 0<=x && x<w && B[y][x]=='.') B[y][x]='@', ans++;
				}
			}
		}
		printf("%d\n",ans);
	}

	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

int main(){
	for(int w,h;scanf("%d%d",&w,&h),w;){
		char B[20][21];
		rep(i,h) scanf("%s",B[i]);

		int ans=1;
		rep(_,h*w){
			rep(i,h) rep(j,w) if(B[i][j]=='@') {
				rep(k,4){
					int y=i+dy[k],x=j+dx[k];
					if(0<=y && y<h && 0<=x && x<w && B[y][x]=='.') B[y][x]='@', ans++;
				}
			}
		}
		printf("%d\n",ans);
	}

	return 0;
}
```