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

4559613
#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;}