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;

int W, H;
char b[20][20];
int ans;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

void dfs(int x, int y) {
	ans++;
	b[y][x] = '#';
	for (int i = 0; i < 4; i++) {
		int nx = x + dx[i], ny = y + dy[i];
		if (0 <= nx && nx < W && 0 <= ny && ny < H && b[ny][nx] == '.') {
			dfs(nx, ny);
		}
	}
}

int main() {
	while (cin >> W >> H , W) {
		for (int y = 0; y < H; y++) {
			for (int x = 0; x < W; x++) {
				cin >> b[y][x];
			}
		}

		ans = 0;

		for (int y = 0; y < H; y++) {
			for (int x = 0; x < W; x++) {
				if (b[y][x] == '@') {
					dfs(x, y);
				}
			}
		}
		cout << ans << endl;
	}
}