#include <cstdio>
using namespace std;

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

int w, h;
char B[20][21];

int dfs(int i, int j) {
   B[i][j] = '#';
   int res = 1;
   for (int k = 0; k < 4; k++) {
      int y = i + dy[k], x = j + dx[k];
      if (y >= 0 && y < h && x >= 0 && x < w && B[y][x] == '.') res += dfs(y, x);
   }
   return res;
}

int main() {
   for (; scanf("%d%d", &w, &h), w;) {
      for (int i = 0; i < h; i++) {
         scanf("%s", B[i]);
      }
      for (int i = 0; i < h; i++) {
         for (int j = 0; j < w; j++) {
            if (B[i][j] == '@') printf("%d\n", dfs(i, j));
         }
      }
   }
}


### Prompt

Create a solution in CPP for 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 <cstdio>
using namespace std;

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

int w, h;
char B[20][21];

int dfs(int i, int j) {
   B[i][j] = '#';
   int res = 1;
   for (int k = 0; k < 4; k++) {
      int y = i + dy[k], x = j + dx[k];
      if (y >= 0 && y < h && x >= 0 && x < w && B[y][x] == '.') res += dfs(y, x);
   }
   return res;
}

int main() {
   for (; scanf("%d%d", &w, &h), w;) {
      for (int i = 0; i < h; i++) {
         scanf("%s", B[i]);
      }
      for (int i = 0; i < h; i++) {
         for (int j = 0; j < w; j++) {
            if (B[i][j] == '@') printf("%d\n", dfs(i, j));
         }
      }
   }
}


```