#include <iostream>
#include <stdio.h>
#include <vector>
using namespace std;

int w, h, ans;
int d[4][2] = { { 1, 0 }, { -1, 0 }, {0, 1}, {0, -1} };
char map[20][20];

void search_dfs(int x, int y){
	if( x < 0 || x >= w || y < 0 || y >= h ) return;
	if( map[y][x] != '.' ) return;
	ans ++;
	map[y][x] = '*';
	for(int i = 0; i < 4; i++)
		search_dfs( x + d[i][0], y + d[i][1] );
}

int main(void){

	while( 1 ) {
		cin >> w >> h;
		if( w == 0 && h == 0 ) break;
		ans = 0;

		for(int i = 0; i < h; i++)
			cin >> map[i];

		for(int i = 0; i < h; i++)
			for(int j = 0; j < w; j++)
				if( map[i][j] == '@' ) {
					map[i][j] = '.';
					search_dfs(j, i);
				}

		cout << ans << endl;
	}

	return 0;
}
### Prompt

Your challenge is to write 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 <stdio.h>
#include <vector>
using namespace std;

int w, h, ans;
int d[4][2] = { { 1, 0 }, { -1, 0 }, {0, 1}, {0, -1} };
char map[20][20];

void search_dfs(int x, int y){
	if( x < 0 || x >= w || y < 0 || y >= h ) return;
	if( map[y][x] != '.' ) return;
	ans ++;
	map[y][x] = '*';
	for(int i = 0; i < 4; i++)
		search_dfs( x + d[i][0], y + d[i][1] );
}

int main(void){

	while( 1 ) {
		cin >> w >> h;
		if( w == 0 && h == 0 ) break;
		ans = 0;

		for(int i = 0; i < h; i++)
			cin >> map[i];

		for(int i = 0; i < h; i++)
			for(int j = 0; j < w; j++)
				if( map[i][j] == '@' ) {
					map[i][j] = '.';
					search_dfs(j, i);
				}

		cout << ans << endl;
	}

	return 0;
}
```