#include<bits/stdc++.h>
using namespace std;

int w,h,ans;
char c[25][25];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

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

int main(int argc, char const *argv[])
{
	while(cin>>w>>h,w)
	{
		ans=0;
		int x,y;
		for (int i = 0; i < h; ++i)
		{
			for (int j = 0; j < w; ++j)
			{
				cin>>c[i][j];
				if(c[i][j]=='@') y=i,x=j,c[i][j]='.';
			}
		}
		dfs(y,x);
		cout<<ans<<endl;
	}
	return 0;
}


### Prompt

Develop a solution in CPP to the problem described below:

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<bits/stdc++.h>
using namespace std;

int w,h,ans;
char c[25][25];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

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

int main(int argc, char const *argv[])
{
	while(cin>>w>>h,w)
	{
		ans=0;
		int x,y;
		for (int i = 0; i < h; ++i)
		{
			for (int j = 0; j < w; ++j)
			{
				cin>>c[i][j];
				if(c[i][j]=='@') y=i,x=j,c[i][j]='.';
			}
		}
		dfs(y,x);
		cout<<ans<<endl;
	}
	return 0;
}


```