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

Solution (in CPP):

Solution:
#include <iostream>
#define rep(i,a,b) for(int i=(a); i<(b); i++)

using namespace std;

int w,h;
char m[22][22];
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};

int dfs(int y , int x){
  int rtn = 0;
  if(y < 0 || h <= y || x < 0 || w <= x || m[y][x] =='#')
    return rtn;
  m[y][x] = '#';
  rtn++;
  rep(i,0,4){
    rtn += dfs(y + dy[i] , x + dx[i]);
  }
  return rtn;
}

int main(){
  while(cin >> w >> h , w + h){
    int sx,sy;
    rep(i,0,h){
      rep(j,0,w){
        cin >> m[i][j];
        if(m[i][j] == '@'){
          sx = j;
          sy = i;
        }
      }
    }
    cout << dfs(sy , sx) << endl;
  }
  return 0;
}