#include<iostream>
#include<queue>
using namespace std;
int w,h;
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
int main()
{
	while(cin>>w>>h,w)
	{
		queue<pair<int,int> >P;
		string s[22];
		int cnt=1;
		bool used[22][22]={};
		for(int i=0;i<h;i++)cin>>s[i];
		for(int i=0;i<h;i++)for(int j=0;j<w;j++)
		{
			if(s[i][j]=='@')
			{
				P.push(make_pair(i,j));
				used[i][j]=1;
			}
		}
		while(!P.empty())
		{
			int x=P.front().first,y=P.front().second;P.pop();
			for(int r=0;r<4;r++)
			{
				int tx=dx[r]+x,ty=y+dy[r];
				if(tx<0||ty<0||tx>=h||ty>=w||s[tx][ty]=='#'||used[tx][ty])continue;
				used[tx][ty]=1;
				P.push(make_pair(tx,ty));
				cnt++;
			}
		}
		cout<<cnt<<endl;
	}
}


### Prompt

Generate 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<queue>
using namespace std;
int w,h;
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
int main()
{
	while(cin>>w>>h,w)
	{
		queue<pair<int,int> >P;
		string s[22];
		int cnt=1;
		bool used[22][22]={};
		for(int i=0;i<h;i++)cin>>s[i];
		for(int i=0;i<h;i++)for(int j=0;j<w;j++)
		{
			if(s[i][j]=='@')
			{
				P.push(make_pair(i,j));
				used[i][j]=1;
			}
		}
		while(!P.empty())
		{
			int x=P.front().first,y=P.front().second;P.pop();
			for(int r=0;r<4;r++)
			{
				int tx=dx[r]+x,ty=y+dy[r];
				if(tx<0||ty<0||tx>=h||ty>=w||s[tx][ty]=='#'||used[tx][ty])continue;
				used[tx][ty]=1;
				P.push(make_pair(tx,ty));
				cnt++;
			}
		}
		cout<<cnt<<endl;
	}
}


```