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
#include<iostream>
using namespace std;

int h,w;
char s[20][20];

void dfs(int y,int x){
	if((x<0)||(w<=x)||(y<0)||(h<=y)){
		return;
	}
	if((s[y][x]=='.')||(s[y][x]=='@')){
		s[y][x]='x';
	}else{
		return;
	}
	dfs(y,x+1);
	dfs(y,x-1);
	dfs(y+1,x);
	dfs(y-1,x);
}

int main(){
	while(cin >> w >> h,h){
		for(int i=0;i<h;i++){
			for(int j=0;j<w;j++){
				cin >> s[i][j];
			}
		}
		for(int i=0;i<h;i++){
			for(int j=0;j<w;j++){
				if(s[i][j]!='@');
				else{
					dfs(i,j);
					j=w;
					i=h;
				}
			}
		}
		int count=0;
		for(int i=0;i<h;i++){
			for(int j=0;j<w;j++){
				if(s[i][j]=='x')
					count++;
			}
		}
		cout << count << endl;
	}
	return 0;
}