#include <iostream>

using namespace std;

int w, h;
char tile[21][21];

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

int dfs(int x, int y){
	if(x<0 || w<=x || y<0 || h<=y) return 0;
	if(tile[y][x] == '#') return 0;

	tile[y][x] = '#';
	int count = 1;
	for(int i=0; i<4; i++){
		count += dfs(x+dx[i], y+dy[i]);
	}
	return count;
}


int main(void){
	while(1){
		cin >> w >> h;
		if(!w && !h) break;

		int x=0, y=0; //pos
		for(int i=0; i<h; i++){
			for(int j=0; j<w; j++){
				cin >> tile[i][j];
				if(tile[i][j] == '@'){
					x = j;
					y = i;
				}
			}
		}
		cout << dfs(x, y) << 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>

using namespace std;

int w, h;
char tile[21][21];

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

int dfs(int x, int y){
	if(x<0 || w<=x || y<0 || h<=y) return 0;
	if(tile[y][x] == '#') return 0;

	tile[y][x] = '#';
	int count = 1;
	for(int i=0; i<4; i++){
		count += dfs(x+dx[i], y+dy[i]);
	}
	return count;
}


int main(void){
	while(1){
		cin >> w >> h;
		if(!w && !h) break;

		int x=0, y=0; //pos
		for(int i=0; i<h; i++){
			for(int j=0; j<w; j++){
				cin >> tile[i][j];
				if(tile[i][j] == '@'){
					x = j;
					y = i;
				}
			}
		}
		cout << dfs(x, y) << endl;
	}
	return 0;
}

```