
#include<iostream>
#include<string>
using namespace std;

string v[20];
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};
int W,H,x,y;
int cnt;

void search(int y, int x)
{
  int i;
  if(v[y][x]=='#') return;
  cnt++;
  v[y][x] = '#';
  for(i=0;i<4;i++){
    if(x+dx[i]<0||W<=x+dx[i]) continue;
    if(y+dy[i]<0||H<=y+dy[i]) continue;
    if(v[y+dy[i]][x+dx[i]] != '.') continue;
    search(y+dy[i],x+dx[i]);
  }
  return;
}

int main()
{
  int i;
  while(cin>>W>>H,W||H){
    for(i=0;i<H;i++){
      cin>>v[i];
      if(v[i].find("@")!=string::npos){
	x=i;
	y=v[i].find("@");
      }
    }
    cnt=0;
    search(x,y);
    cout<<cnt<<endl;
  }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;

string v[20];
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};
int W,H,x,y;
int cnt;

void search(int y, int x)
{
  int i;
  if(v[y][x]=='#') return;
  cnt++;
  v[y][x] = '#';
  for(i=0;i<4;i++){
    if(x+dx[i]<0||W<=x+dx[i]) continue;
    if(y+dy[i]<0||H<=y+dy[i]) continue;
    if(v[y+dy[i]][x+dx[i]] != '.') continue;
    search(y+dy[i],x+dx[i]);
  }
  return;
}

int main()
{
  int i;
  while(cin>>W>>H,W||H){
    for(i=0;i<H;i++){
      cin>>v[i];
      if(v[i].find("@")!=string::npos){
	x=i;
	y=v[i].find("@");
      }
    }
    cnt=0;
    search(x,y);
    cout<<cnt<<endl;
  }
}
```