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:
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
#define EPS 1e-10
using namespace std;
typedef long long llong;
vector<vector<char> >v;
int w, h;
int ans = 0;
void dfs(int x, int y) {
	if (x < 0 || h <= x)return;
	if (y < 0 || w <= y)return;

	if (v[x][y] == '#')return;

	if (v[x][y] == '.'||v[x][y]=='@') {
		ans++;
		v[x][y] = '#';
		dfs(x + 1, y);
		dfs(x - 1, y);
		dfs(x, y - 1);
		dfs(x, y + 1);
	}
	return;
}
int main() {
	while (cin >> w >> h, h) {
		ans = 0;
		int x, y;
		v = vector<vector<char> >(h, vector<char>(w));
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				cin >> v[i][j];
				if (v[i][j] == '@') {
					x = i;
					y = j;
				}
			}
		}
		dfs(x, y);
		cout << ans << endl;
	}
	return 0;
}