//00
#include<iostream>
#include<queue>
#include<utility>

using namespace std;

int main(){
  for(int W,H;cin>>W>>H,W|H;){
    bool g[20][20];
    queue<pair<int,int> > que;
    for(int y=0;y<H;y++){
      for(int x=0;x<W;x++){
	char c;
	cin>>c;
	g[y][x]=c=='.';
	if(c=='@'){
	  que.push(make_pair(y,x));
	}
      }
    }
    int ans=0;
    while(!que.empty()){
      ans++;
      int y=que.front().first;
      int x=que.front().second;
      que.pop();
      int dy[]={0,1,0,-1};
      int dx[]={1,0,-1,0};
      for(int i=0;i<4;i++){
	int ny=y+dy[i];
	int nx=x+dx[i];
	if(0<=ny&&ny<H&&0<=nx&&nx<W&&g[ny][nx]){
	  que.push(make_pair(ny,nx));
	  g[ny][nx]=false;
	}
      }
    }
    cout<<ans<<endl;
  }
  return 0;
}
### 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
//00
#include<iostream>
#include<queue>
#include<utility>

using namespace std;

int main(){
  for(int W,H;cin>>W>>H,W|H;){
    bool g[20][20];
    queue<pair<int,int> > que;
    for(int y=0;y<H;y++){
      for(int x=0;x<W;x++){
	char c;
	cin>>c;
	g[y][x]=c=='.';
	if(c=='@'){
	  que.push(make_pair(y,x));
	}
      }
    }
    int ans=0;
    while(!que.empty()){
      ans++;
      int y=que.front().first;
      int x=que.front().second;
      que.pop();
      int dy[]={0,1,0,-1};
      int dx[]={1,0,-1,0};
      for(int i=0;i<4;i++){
	int ny=y+dy[i];
	int nx=x+dx[i];
	if(0<=ny&&ny<H&&0<=nx&&nx<W&&g[ny][nx]){
	  que.push(make_pair(ny,nx));
	  g[ny][nx]=false;
	}
      }
    }
    cout<<ans<<endl;
  }
  return 0;
}
```