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<cstring>
typedef long long ll;
typedef long double ld;
using namespace std;
ll dx[] = {1, -1, 0, 0};
ll dy[] = { 0,0,1,-1 };
ll W, H;
ll cnt;
char field[100][100];
void dfs(ll y, ll x) {
	field[y][x] = '#';
	for (ll k = 0; k < 4; k++) {
		ll ny = y + dy[k];
		ll nx = x + dx[k];
		if (ny >= 0 && ny <H && nx >= 0 && nx < W) {
			if (field[ny][nx] == '.')
			{
				cnt++;
				dfs(ny, nx);
			}
		}
	}
	return;
}
int main() {
	while (cin >> W >> H) {
		if (W == 0 && H == 0) { break; }
		cnt = 1;
		for (ll i = 0; i < H; i++) {
			cin >> field[i];
		}
		for (ll i = 0; i < H; i++) {
			for (ll j = 0; j < W; j++) {
				if (field[i][j] == '@') {
					dfs(i, j);
				}
			}
		}
		cout << cnt << endl;
	}
}