#include<iostream>
using namespace std;

int ans,w,h;
string g[30];
bool visit[30][30];

int rec(int y,int x){
  int res = 1;
  int dx[] = {0,1,0,-1}, dy[] = {-1,0,1,0};
  visit[y][x] = true;

  for(int k=0;k<4;k++){
    int sy = y+dy[k] ,sx = x+dx[k];
    if(sy<0 || sx<0 || h<=sy || w<=sx)continue;
    if(visit[sy][sx] || g[sy][sx] == '#')continue;
    res += rec(sy,sx);
  }
  return res;
}

int main(){
  while(cin >> w >> h && (w||h)){

    for(int i=0;i<h;i++)for(int j=0;j<w;j++)visit[i][j] = false;
    for(int i=0;i<h;i++)cin >> g[i];

    ans = 0;
    for(int i=0;i<h;i++){
      for(int j=0;j<w;j++){
	if(g[i][j] == '@')cout << rec(i,j) << endl;
      }
    }
  }
}
### Prompt

Your challenge is to write 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>
using namespace std;

int ans,w,h;
string g[30];
bool visit[30][30];

int rec(int y,int x){
  int res = 1;
  int dx[] = {0,1,0,-1}, dy[] = {-1,0,1,0};
  visit[y][x] = true;

  for(int k=0;k<4;k++){
    int sy = y+dy[k] ,sx = x+dx[k];
    if(sy<0 || sx<0 || h<=sy || w<=sx)continue;
    if(visit[sy][sx] || g[sy][sx] == '#')continue;
    res += rec(sy,sx);
  }
  return res;
}

int main(){
  while(cin >> w >> h && (w||h)){

    for(int i=0;i<h;i++)for(int j=0;j<w;j++)visit[i][j] = false;
    for(int i=0;i<h;i++)cin >> g[i];

    ans = 0;
    for(int i=0;i<h;i++){
      for(int j=0;j<w;j++){
	if(g[i][j] == '@')cout << rec(i,j) << endl;
      }
    }
  }
}
```