#include <bits/stdc++.h>
using namespace std;

const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
int W, H;
char board[30][30];

int dfs(int y, int x) {
   board[y][x] = '#';
   int ret = 1;
   for (int i = 0; i < 4; i++) {
      int nx = x + dx[i], ny = y + dy[i];
      if (nx >= 0 && nx < W && ny >= 0 && ny < H && board[ny][nx] == '.') {
         ret += dfs(ny, nx);
      }
   }
   return ret;
}

int main() {
   while (cin >> W >> H, W) {
      for (int i = 0; i < H; i++) {
         cin >> board[i];
      }
      for (int i = 0; i < H; i++) {
         for (int j = 0; j < W; j++) {
            if (board[i][j] == '@') {
               cout << dfs(i, j) << endl;
            }
         }
      }
   }

   return 0;
}

### Prompt

Please provide a CPP coded solution 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 <bits/stdc++.h>
using namespace std;

const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
int W, H;
char board[30][30];

int dfs(int y, int x) {
   board[y][x] = '#';
   int ret = 1;
   for (int i = 0; i < 4; i++) {
      int nx = x + dx[i], ny = y + dy[i];
      if (nx >= 0 && nx < W && ny >= 0 && ny < H && board[ny][nx] == '.') {
         ret += dfs(ny, nx);
      }
   }
   return ret;
}

int main() {
   while (cin >> W >> H, W) {
      for (int i = 0; i < H; i++) {
         cin >> board[i];
      }
      for (int i = 0; i < H; i++) {
         for (int j = 0; j < W; j++) {
            if (board[i][j] == '@') {
               cout << dfs(i, j) << endl;
            }
         }
      }
   }

   return 0;
}

```