#include <iostream>
#include <vector>
#include <string>

using namespace std;

int w, h;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
int dfs(vector<string>& mp, int x, int y){
  int cnt = 0;  
  for(int i=0;i<4;++i){
	int tx = x + dx[i];
	int ty = y + dy[i];
	if(tx < 0 || w <= tx || ty < 0 || h <= ty) continue;
	if(mp[ty][tx] == '#') continue;
	mp[ty][tx] = '#';
	cnt += dfs(mp, tx, ty) + 1;
  }
  return cnt;
}

int main(){
  while(true){
	if(!(cin >> w >> h)) return 0;
	if(w == 0 && h == 0) break;
	vector<string> vs(h);
	for(int i=0;i<h;++i)
	  cin >> vs[i];
	int x, y;
	for(int i=0;i<h;++i)
	  for(int j=0;j<w;++j)
		if(vs[i][j] == '@') x=j, y=i;
	vs[y][x] = '#';
	cout << 1+dfs(vs,x,y) << endl;
  }
  return 0;
}
### Prompt

Create a solution in cpp for 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 <vector>
#include <string>

using namespace std;

int w, h;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
int dfs(vector<string>& mp, int x, int y){
  int cnt = 0;  
  for(int i=0;i<4;++i){
	int tx = x + dx[i];
	int ty = y + dy[i];
	if(tx < 0 || w <= tx || ty < 0 || h <= ty) continue;
	if(mp[ty][tx] == '#') continue;
	mp[ty][tx] = '#';
	cnt += dfs(mp, tx, ty) + 1;
  }
  return cnt;
}

int main(){
  while(true){
	if(!(cin >> w >> h)) return 0;
	if(w == 0 && h == 0) break;
	vector<string> vs(h);
	for(int i=0;i<h;++i)
	  cin >> vs[i];
	int x, y;
	for(int i=0;i<h;++i)
	  for(int j=0;j<w;++j)
		if(vs[i][j] == '@') x=j, y=i;
	vs[y][x] = '#';
	cout << 1+dfs(vs,x,y) << endl;
  }
  return 0;
}
```