#include <iostream>
using namespace std;

int h, w, ans;
char field[25][25];

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

void dfs(int x, int y) {
  if(field[y][x] == '.' && 0 <= x && x < w && 0 <= y && y < h) {
    field[y][x] = '#';
    ans++;
    for(int i = 0; i < 4; i++) {
      int nx = x+dx[i], ny = y+dy[i];
      dfs(nx, ny);
    }
  }
}

int main() {
  int sx, sy;
  while(cin >> w >> h, w+h) {
    for(int i = 0; i < h; i++) {
      for(int j = 0; j < w; j++) {
	cin >> field[i][j];
	if(field[i][j] == '@') {
	  sx = j;
	  sy = i;
	  field[i][j] = '.';
	}
      }
    }

    ans = 0;
    dfs(sx, sy);
    cout << ans << endl;
  }
}
### Prompt

In cpp, your task is to solve 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>
using namespace std;

int h, w, ans;
char field[25][25];

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

void dfs(int x, int y) {
  if(field[y][x] == '.' && 0 <= x && x < w && 0 <= y && y < h) {
    field[y][x] = '#';
    ans++;
    for(int i = 0; i < 4; i++) {
      int nx = x+dx[i], ny = y+dy[i];
      dfs(nx, ny);
    }
  }
}

int main() {
  int sx, sy;
  while(cin >> w >> h, w+h) {
    for(int i = 0; i < h; i++) {
      for(int j = 0; j < w; j++) {
	cin >> field[i][j];
	if(field[i][j] == '@') {
	  sx = j;
	  sy = i;
	  field[i][j] = '.';
	}
      }
    }

    ans = 0;
    dfs(sx, sy);
    cout << ans << endl;
  }
}
```