#include<iostream>
using namespace std;
#define MAX (256)

char field[MAX][MAX];
int W, H;
int ans = 0;
int sx, sy;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

void dfs(int x, int y){
	if(0 <= x && x < H && 0 <= y && y < W && field[x][y] == '.'){
		ans++;
		field[x][y] = '#';

		for(int i=0; i < 4; i++){
			int nx = x + dx[i]; int ny = y + dy[i];
			dfs(nx, ny);
		}
	}
}

int main(){
	while(cin >> W >> H, W || H){
		ans = 0;
		for(int i=0; i < H; i++){
			for(int j=0; j < W; j++){
				cin >> field[i][j];
				if(field[i][j] == '@'){
					sx = i;
					sy = j;
					field[i][j] = '.';
				}
			}
		}
		dfs(sx, sy);

		cout << ans << endl;
	}

}
### 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>
using namespace std;
#define MAX (256)

char field[MAX][MAX];
int W, H;
int ans = 0;
int sx, sy;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

void dfs(int x, int y){
	if(0 <= x && x < H && 0 <= y && y < W && field[x][y] == '.'){
		ans++;
		field[x][y] = '#';

		for(int i=0; i < 4; i++){
			int nx = x + dx[i]; int ny = y + dy[i];
			dfs(nx, ny);
		}
	}
}

int main(){
	while(cin >> W >> H, W || H){
		ans = 0;
		for(int i=0; i < H; i++){
			for(int j=0; j < W; j++){
				cin >> field[i][j];
				if(field[i][j] == '@'){
					sx = i;
					sy = j;
					field[i][j] = '.';
				}
			}
		}
		dfs(sx, sy);

		cout << ans << endl;
	}

}
```