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 <vector>

using namespace std;

vector<string> v;
int w, h;

int dfs(int i, int j) {
	int cnt = 0;
	if (i < 0 || j < 0 || i >= h || j >= w) return 0;
	if (v[i][j] == '#') return 0;
	v[i][j] = '#';
	cnt++;
	cnt += dfs(i, j - 1) + dfs(i, j + 1) + dfs(i - 1, j) + dfs(i + 1, j);
	return cnt;
}

int main()
{
	int di, dj;

	while (cin >> w >> h, w) {
		v = vector<string>(h);
		for (int i = 0; i < h; i++) {
			cin >> v[i];
			for (int j = 0; j < w; j++) {	
				if (v[i][j] == '@') {
					di = i;
					dj = j;
				} 
			}
		}
		cout << dfs(di, dj) << endl;
	}
	return 0;
}