#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <iterator>

using namespace std;

int w, h;
string str[30];

int search(int y, int x){
	if(x < 0 || x >= w || y < 0 || y >= h){
		return 0;
	}
	if(str[y][x] == '#'){
		return 0;
	}
	str[y][x] = '#';
	int ret = 1;
	ret += search(y, x - 1);
	ret += search(y, x + 1);
	ret += search(y - 1, x);
	ret += search(y + 1, x);
	return ret;
}

int main(){
	
	while(true){
		cin >> w >> h;
		if(w == 0 && h == 0){
			break;
		}
		for(int i = 0; i < h; i++){
			cin >> str[i];
		}
		int ans;
		for(int i = 0; i < h; i++){
			for(int j = 0; j < w; j++){
				if(str[i][j] == '@'){
					ans = search(i, j);
				}
			}
		}
		cout << ans << endl;
	}
	
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <iterator>

using namespace std;

int w, h;
string str[30];

int search(int y, int x){
	if(x < 0 || x >= w || y < 0 || y >= h){
		return 0;
	}
	if(str[y][x] == '#'){
		return 0;
	}
	str[y][x] = '#';
	int ret = 1;
	ret += search(y, x - 1);
	ret += search(y, x + 1);
	ret += search(y - 1, x);
	ret += search(y + 1, x);
	return ret;
}

int main(){
	
	while(true){
		cin >> w >> h;
		if(w == 0 && h == 0){
			break;
		}
		for(int i = 0; i < h; i++){
			cin >> str[i];
		}
		int ans;
		for(int i = 0; i < h; i++){
			for(int j = 0; j < w; j++){
				if(str[i][j] == '@'){
					ans = search(i, j);
				}
			}
		}
		cout << ans << endl;
	}
	
	return 0;
}
```