Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstring>
using namespace std;

int w,h;
char m[22][32];

int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
int dfs(int y,int x){
	if(m[y][x]=='#')return 0;
	int res=1;
	m[y][x]='#';
	for(int i=0;i<4;i++){
		res+=dfs(y+dy[i],x+dx[i]);
	}
	return res;
}
int main(){
	while(cin>>w>>h,w|h){
		memset(m,'#',sizeof(m));
		for(int i=1;i<=h;i++)for(int j=1;j<=w;j++)cin>>m[i][j];
		for(int i=1;i<=h;i++){
			for(int j=1;j<=w;j++){
				if(m[i][j]!='@')continue;
				cout<<dfs(i,j)<<endl;
				goto BREAK;
			}
		}
		BREAK:;
	}
	return 0;
}