#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
#define REP(i, j) for(int i = 0; i < j; i++)

int w, h;

int dfs(int y, int x, vector<string> &v){
  int my[] = {0, 0, 1, -1};
  int mx[] = {1, -1, 0, 0};
  int ret = 1;
  REP(i, 4){
    int ny = y + my[i], nx = x + mx[i];
    if(ny >= 0 && nx >= 0 && ny < h && nx < w && v[ny][nx] == '.'){
      v[ny][nx] = '#';
      ret += dfs(ny, nx, v);
    }
  }
  return ret;
}

int main(){
  while(cin >>w >>h && w){
    vector<string> v(h);
    REP(i, h) cin >>v[i];
    REP(i, h) REP(j, w) if(v[i][j] == '@') cout <<dfs(i, j, v) <<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<vector>
#include<string>
#include<algorithm>
using namespace std;
#define REP(i, j) for(int i = 0; i < j; i++)

int w, h;

int dfs(int y, int x, vector<string> &v){
  int my[] = {0, 0, 1, -1};
  int mx[] = {1, -1, 0, 0};
  int ret = 1;
  REP(i, 4){
    int ny = y + my[i], nx = x + mx[i];
    if(ny >= 0 && nx >= 0 && ny < h && nx < w && v[ny][nx] == '.'){
      v[ny][nx] = '#';
      ret += dfs(ny, nx, v);
    }
  }
  return ret;
}

int main(){
  while(cin >>w >>h && w){
    vector<string> v(h);
    REP(i, h) cin >>v[i];
    REP(i, h) REP(j, w) if(v[i][j] == '@') cout <<dfs(i, j, v) <<endl;
  }
  return 0;
}
```