#include<iostream>
#include<string>

#define rep(i,n) for(int i=0; i<n; i++)

std::string field[20];
int dx[4] = {0,-1,0,1};	//右、上、左、下
int dy[4] = {1,0,-1,0};
int h,w;
int cnt;
void rec(int y,int x){
	field[y][x] = '#';
	cnt++;
	int xi, yi;
	rep(i,4){
		xi = x, yi = y;
		xi += dx[i];
		yi += dy[i];

		if(xi < 0 || xi >= w)continue;
		if(yi < 0 || yi >= h)continue;
		if(field[yi][xi] != '.')continue;
		rec(yi,xi);
	}
}

int main(){

	while(std::cin >> w >> h, (h||w)){

		cnt = 0;

		rep(i,h){
			std::cin >> field[i];
		}
		rep(i,h){
			rep(j,w){
				if(field[i][j] == '@'){
					rec(i,j);
				}
			}
		}
		std::cout << cnt << std::endl;
	}
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<string>

#define rep(i,n) for(int i=0; i<n; i++)

std::string field[20];
int dx[4] = {0,-1,0,1};	//右、上、左、下
int dy[4] = {1,0,-1,0};
int h,w;
int cnt;
void rec(int y,int x){
	field[y][x] = '#';
	cnt++;
	int xi, yi;
	rep(i,4){
		xi = x, yi = y;
		xi += dx[i];
		yi += dy[i];

		if(xi < 0 || xi >= w)continue;
		if(yi < 0 || yi >= h)continue;
		if(field[yi][xi] != '.')continue;
		rec(yi,xi);
	}
}

int main(){

	while(std::cin >> w >> h, (h||w)){

		cnt = 0;

		rep(i,h){
			std::cin >> field[i];
		}
		rep(i,h){
			rep(j,w){
				if(field[i][j] == '@'){
					rec(i,j);
				}
			}
		}
		std::cout << cnt << std::endl;
	}
	return 0;
}
```