#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cstdio>
using namespace std;


int W, H;
char Map[25][25];
int d[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int dfs(int x, int y){
	if(Map[y][x] == '#') return 0;
	Map[y][x] = '#';
	int ret = 1;
	for(int i=0;i<4;i++){
		ret += dfs(x+d[i][1],y+d[i][0]);
	}
	return ret;
}
int main(){
	int i,j;
	while(cin >> W >> H, W){
		int sx, sy;
		for(i=0;i<=H+1;i++)
			for(j=0;j<=W+1;j++)
				Map[i][j] = '#';
		for(i=1;i<=H;i++){
			for(j=1;j<=W;j++){
				cin >> Map[i][j];
				if(Map[i][j] == '@'){
					sx = j;sy = i;
				}
			}
		}
		cout << dfs(sx, sy) << 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 <algorithm>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cstdio>
using namespace std;


int W, H;
char Map[25][25];
int d[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int dfs(int x, int y){
	if(Map[y][x] == '#') return 0;
	Map[y][x] = '#';
	int ret = 1;
	for(int i=0;i<4;i++){
		ret += dfs(x+d[i][1],y+d[i][0]);
	}
	return ret;
}
int main(){
	int i,j;
	while(cin >> W >> H, W){
		int sx, sy;
		for(i=0;i<=H+1;i++)
			for(j=0;j<=W+1;j++)
				Map[i][j] = '#';
		for(i=1;i<=H;i++){
			for(j=1;j<=W;j++){
				cin >> Map[i][j];
				if(Map[i][j] == '@'){
					sx = j;sy = i;
				}
			}
		}
		cout << dfs(sx, sy) << endl;
	}
	return 0;
}
```