#include <iostream>
using namespace std;
#define rep(i, n) for (int i = 0; i < int(n); ++i)

char table[20][20];
	int W,H;

void dfs(int x,int y){
	if(table[x][y]!= '#')
		table[x][y]='#';
	int dx[4]={1,0,-1,0};
	int dy[4]={0,1,0,-1};
	rep(i,4)
		if(0<=x+dx[i]&&x+dx[i]<W&&0<=y+dy[i]&&y+dy[i]<H&&table[x+dx[i]][y+dy[i]]!='#')
			dfs(x+dx[i],y+dy[i]);
	return ;
}


int main(void){
	while(cin >> W >> H,W|H){
		int pre=0,sx,sy;
		rep(y,H){
			rep(x,W){
				cin >> table[x][y];
				if(table[x][y]=='#')
					pre++;
				if(table[x][y]=='@')
					sx=x,sy=y;
			}
		}
		dfs(sx,sy);
		int post=0;
		rep(y,H)
			rep(x,W)
				if(table[x][y]=='#')
					post++;
		cout << post-pre << endl;
	}
	return 0;
}
### Prompt

Develop a solution in cpp 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 <iostream>
using namespace std;
#define rep(i, n) for (int i = 0; i < int(n); ++i)

char table[20][20];
	int W,H;

void dfs(int x,int y){
	if(table[x][y]!= '#')
		table[x][y]='#';
	int dx[4]={1,0,-1,0};
	int dy[4]={0,1,0,-1};
	rep(i,4)
		if(0<=x+dx[i]&&x+dx[i]<W&&0<=y+dy[i]&&y+dy[i]<H&&table[x+dx[i]][y+dy[i]]!='#')
			dfs(x+dx[i],y+dy[i]);
	return ;
}


int main(void){
	while(cin >> W >> H,W|H){
		int pre=0,sx,sy;
		rep(y,H){
			rep(x,W){
				cin >> table[x][y];
				if(table[x][y]=='#')
					pre++;
				if(table[x][y]=='@')
					sx=x,sy=y;
			}
		}
		dfs(sx,sy);
		int post=0;
		rep(y,H)
			rep(x,W)
				if(table[x][y]=='#')
					post++;
		cout << post-pre << endl;
	}
	return 0;
}
```