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>
using namespace std;
char m[20][21];
int h, w, n;
void dfs(int x, int y)
{
	int i, j, nx, ny, dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
	m[y][x] = '#';
	n++;
	for (i = 0; i < 4; i++)
	{
		nx = x + dx[i];
		ny = y + dy[i];
		if (0 <= nx && nx < w && 0 <= ny && ny < h && m[ny][nx] == '.') dfs(nx, ny);
	}
	return;
}
int main()
{
	int i, j;
	while (cin >> w >> h)
	{
		if (h == 0 && w == 0) break;
		for (i = 0; i < h; i++)
		{
			cin >> m[i];
			if (cin.eof()) return 0;
		}
		n = 0;
		for (i = 0; i < h; i++)
		{
			for (j = 0; j < w; j++)
			{
				if (m[i][j] == '@')
				{
					dfs(j, i);
				}
			}
		}
		cout << n << endl;
	}
}