Red and Black

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) 

 

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

 

#include<iostream>#include<cmath>#include<algorithm>using namespace std;char a[100][100];bool vis[100][100];int ans;int n,k;void dfs(int x,int y){    if(x-1>=0&&x<k&&y>=0&&y<n&&a[x-1][y]=='.'&&!vis[x-1][y])    {        vis[x-1][y]=true;        ans++;        dfs(x-1,y);    }    if(x>=0&&x+1<k&&y>=0&&y<n&&a[x+1][y]=='.'&&!vis[x+1][y])    {        vis[x+1][y]=true;        ans++;        dfs(x+1,y);    }    if(x>=0&&x<k&&y-1>=0&&y<n&&a[x][y-1]=='.'&&!vis[x][y-1])    {        vis[x][y-1]=true;        ans++;        dfs(x,y-1);    }    if(x>=0&&x<k&&y>=0&&y+1<n&&a[x][y+1]=='.'&&!vis[x][y+1])    {        vis[x][y+1]=true;        ans++;        dfs(x,y+1);    }}int main(){    while(cin>>n>>k,n!=0&&k!=0)    {        int sx,sy;        memset(vis,0,sizeof(vis));        for(int i=0;i<k;i++)            for(int j=0;j<n;j++)            {                cin>>a[i][j];                if(a[i][j]=='@')                {                    sx=i;                    sy=j;                    vis[i][j]=true;                }            }            ans=1;            dfs(sx,sy);            cout<<ans<<"\n";    }}