#include<iostream>
#include<string>

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

int w,h;
std::string f[20];
int cnt;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,-1,0,1};

void rec(int x, int y){
	int mx, my;

	f[y][x] = '#';
	cnt++;

	rep(i,4){
		mx = x + dx[i];
		my = y + dy[i];

		if(mx < 0 || mx >= w)continue;
		if(my < 0 || my >= h)continue;
		if(f[my][mx] != '.')continue;
		rec(mx, my);
	}
}




int main(){
	while(std::cin >> w >> h, (w||h)){
		rep(i,h){
			std::cin >> f[i];
		}
		cnt = 0;
		rep(i,h){
			rep(j,w){
				if(f[i][j] == '@')rec(j,i);
			}
		}
		std::cout << cnt << std::endl;
	}
	return 0;
}
### Prompt

Please create a solution in Cpp 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>
#include<string>

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

int w,h;
std::string f[20];
int cnt;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,-1,0,1};

void rec(int x, int y){
	int mx, my;

	f[y][x] = '#';
	cnt++;

	rep(i,4){
		mx = x + dx[i];
		my = y + dy[i];

		if(mx < 0 || mx >= w)continue;
		if(my < 0 || my >= h)continue;
		if(f[my][mx] != '.')continue;
		rec(mx, my);
	}
}




int main(){
	while(std::cin >> w >> h, (w||h)){
		rep(i,h){
			std::cin >> f[i];
		}
		cnt = 0;
		rep(i,h){
			rep(j,w){
				if(f[i][j] == '@')rec(j,i);
			}
		}
		std::cout << cnt << std::endl;
	}
	return 0;
}
```