#include <iostream>
#include <string>

using namespace std;
int W, H;
string s[20];
int num;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {-1, 0, 1, 0};

void dfs(int x, int y){
  num++;
  s[y][x] = '#';
  for(int i=0; i<4; i++){
    if(0<=x+dx[i] && x+dx[i]<W && 0<=y+dy[i] && y+dy[i]<H && s[y+dy[i]][x+dx[i]]=='.'){
      dfs(x+dx[i], y+dy[i]);
    }
  }
}

int main(){
  while(cin >> W >> H, W || H){
    for(int i=0; i<H; i++){
      cin >> s[i];
    }
    int x, y;
    for(int i=0; i<H; i++){
      for(int j=0; j<W; j++){
	if(s[i][j] == '@'){
	  x = j;
	  y = i;
	}
      }
    }
    num = 0;
    dfs(x, y);
    cout << num << 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>

using namespace std;
int W, H;
string s[20];
int num;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {-1, 0, 1, 0};

void dfs(int x, int y){
  num++;
  s[y][x] = '#';
  for(int i=0; i<4; i++){
    if(0<=x+dx[i] && x+dx[i]<W && 0<=y+dy[i] && y+dy[i]<H && s[y+dy[i]][x+dx[i]]=='.'){
      dfs(x+dx[i], y+dy[i]);
    }
  }
}

int main(){
  while(cin >> W >> H, W || H){
    for(int i=0; i<H; i++){
      cin >> s[i];
    }
    int x, y;
    for(int i=0; i<H; i++){
      for(int j=0; j<W; j++){
	if(s[i][j] == '@'){
	  x = j;
	  y = i;
	}
      }
    }
    num = 0;
    dfs(x, y);
    cout << num << endl;
  }
  return 0;
}
```