#include<iostream>
#include<cstdio>
#include<string.h>
#define rep(i,n) for(int i =0;i<n;i++)
using namespace std;

string t[21];
int w,h;

void dfs(int x,int y){
int dx[4] = {1,-1, 0, 0};
int dy[4] = {0, 0, 1, -1};


	if(((x<0||y<0)||x>=h||y>=w)||t[x][y]!='.') return;
	t[x][y]='@';	
	rep(k,4){
	dfs(x+dx[k],y+dy[k]);
}
return;
}


int main(void){
	int sx=-1,sy=-1,cnt;
	
while(true){
scanf("%d %d/n",&w,&h);
if(!w&&!h) {break;}
rep(i,h){
cin >> t[i];
}

rep(i,h)rep(j,w){

	if(t[i][j]=='@'){	
	sx =i;
	sy =j;

t[i][j]='.';
	}

}

dfs(sx,sy);

cnt =0;

rep(i,h)rep(j,w)if(t[i][j]=='@') cnt++;	

cout<<cnt<<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<iostream>
#include<cstdio>
#include<string.h>
#define rep(i,n) for(int i =0;i<n;i++)
using namespace std;

string t[21];
int w,h;

void dfs(int x,int y){
int dx[4] = {1,-1, 0, 0};
int dy[4] = {0, 0, 1, -1};


	if(((x<0||y<0)||x>=h||y>=w)||t[x][y]!='.') return;
	t[x][y]='@';	
	rep(k,4){
	dfs(x+dx[k],y+dy[k]);
}
return;
}


int main(void){
	int sx=-1,sy=-1,cnt;
	
while(true){
scanf("%d %d/n",&w,&h);
if(!w&&!h) {break;}
rep(i,h){
cin >> t[i];
}

rep(i,h)rep(j,w){

	if(t[i][j]=='@'){	
	sx =i;
	sy =j;

t[i][j]='.';
	}

}

dfs(sx,sy);

cnt =0;

rep(i,h)rep(j,w)if(t[i][j]=='@') cnt++;	

cout<<cnt<<endl;;


}

return 0;
}
```