#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int h,w;
char grid[21][21];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};

int dfs(int x,int y){
  if(grid[y][x]=='#')return 0;
  grid[y][x]='#';
  int sum=1;
  for(int i=0;i<4;i++){
    int nx=x+dx[i],ny=y+dy[i];
    if(0<=nx && nx<w && 0<=ny && ny<h)sum+=dfs(nx,ny);
  }
  return sum;
}

int main(void){

  while(cin >> w >> h,w|h){
    for(int i=0;i<h;i++)cin >> grid[i];
    
    for(int y=0;y<h;y++)
      for(int x=0;x<w;x++)
	if(grid[y][x]=='@')cout << dfs(x,y) << endl;
  }

  return 0;
}
### Prompt

Please formulate a cpp solution 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<vector>
#include<algorithm>

using namespace std;

int h,w;
char grid[21][21];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};

int dfs(int x,int y){
  if(grid[y][x]=='#')return 0;
  grid[y][x]='#';
  int sum=1;
  for(int i=0;i<4;i++){
    int nx=x+dx[i],ny=y+dy[i];
    if(0<=nx && nx<w && 0<=ny && ny<h)sum+=dfs(nx,ny);
  }
  return sum;
}

int main(void){

  while(cin >> w >> h,w|h){
    for(int i=0;i<h;i++)cin >> grid[i];
    
    for(int y=0;y<h;y++)
      for(int x=0;x<w;x++)
	if(grid[y][x]=='@')cout << dfs(x,y) << endl;
  }

  return 0;
}
```