#include <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)

int dx[4] = {0,1,0,-1}, dy[4] = {1,0,-1,0};

int W, H, ans;
string grid[22];

int rec(int x, int y) {
	int res = 1;
	
	grid[y][x] = '#';
	
	for_(d,0,4) {
		int nx = x + dx[d], ny = y + dy[d];
		if (nx < 0 || W <= nx || ny < 0 || H <= ny) continue;
		if (grid[ny][nx] == '#') continue;
		res += rec(nx, ny);
	}
	
	return res;
}

int main() {
	while (cin >> W >> H, W) {
		for_(i,0,H) cin >> grid[i];
		
		for_(i,0,H) for_(j,0,W) if (grid[i][j] == '@') cout << rec(j, i) << endl;
	}
}
### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)

int dx[4] = {0,1,0,-1}, dy[4] = {1,0,-1,0};

int W, H, ans;
string grid[22];

int rec(int x, int y) {
	int res = 1;
	
	grid[y][x] = '#';
	
	for_(d,0,4) {
		int nx = x + dx[d], ny = y + dy[d];
		if (nx < 0 || W <= nx || ny < 0 || H <= ny) continue;
		if (grid[ny][nx] == '#') continue;
		res += rec(nx, ny);
	}
	
	return res;
}

int main() {
	while (cin >> W >> H, W) {
		for_(i,0,H) cin >> grid[i];
		
		for_(i,0,H) for_(j,0,W) if (grid[i][j] == '@') cout << rec(j, i) << endl;
	}
}
```