Red and Black

Description

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).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13



#include <iostream>
using namespace std;
char a[100][100];

int p(char a[100][100],int x,int y)
{int l=0;
if(a[x-1][y]=='.')
{a[x-1][y]='1';p(a,x-1,y);l=1;}
if(a[x+1][y]=='.')
{a[x+1][y]='1';p(a,x+1,y);l=1;}
if(a[x][y-1]=='.')
{a[x][y-1]='1';p(a,x,y-1);l=1;}
if(a[x][y+1]=='.')
{a[x][y+1]='1';p(a,x,y+1);l=1;}
if(l==0)
return 0;
}


int main()
{int a1,a2,b1,b2,i,j;
while(cin>>a2>>a1,a2!=0)
{for(i=1;i<=a1;i++)
for(j=1;j<=a2;j++)
{cin>>a[i][j];
if(a[i][j]=='@')
{b1=i;b2=j;}
}
for(i=0;i<=a1;i++)
{a[i][a2+1]='#';
a[i][0]='#';}
for(i=0;i<=a2;i++)
{a[a1+1][i]='#';
a[0][i]='#';}
p(a,b1,b2);
int count=0;
for(i=1;i<=a1;i++)
for(j=1;j<=a2;j++)
{if(a[i][j]=='1')
count++;}

cout<<count+1<<endl;


}

return 0;}