#include <iostream>
using namespace std;

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

int w,h;
char f[32][32];

int fill(int x, int y){
	if( x<0 || y<0 || x>=w || y>=h || f[y][x]=='#' )return 0;
	f[y][x] = '#';
	return 1 + fill(x-1,y) + fill(x+1,y) + fill(x,y+1) + fill(x,y-1);
}

int main(){
	int x,y;
	while(cin>>w>>h,w|h){
		rep(i,h)cin>>f[i];
		rep(i,h)rep(j,w)if( f[i][j]=='@' ){
			x=j; y=i;
		}
		printf("%d\n",fill(x,y));
	}
	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 <iostream>
using namespace std;

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

int w,h;
char f[32][32];

int fill(int x, int y){
	if( x<0 || y<0 || x>=w || y>=h || f[y][x]=='#' )return 0;
	f[y][x] = '#';
	return 1 + fill(x-1,y) + fill(x+1,y) + fill(x,y+1) + fill(x,y-1);
}

int main(){
	int x,y;
	while(cin>>w>>h,w|h){
		rep(i,h)cin>>f[i];
		rep(i,h)rep(j,w)if( f[i][j]=='@' ){
			x=j; y=i;
		}
		printf("%d\n",fill(x,y));
	}
	return 0;
}
```