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>
#include<string>

using namespace std;

string a[21];
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int cnt, X, Y;

void dfs(int y, int x){
  if( y < 0 || Y <= y || x < 0 || X <= x || a[y][x] == '#') return;
  a[y][x] = '#';
  cnt++;
  for(int i = 0; i < 4; i++)
    dfs(y + dy[i], x + dx[i]);
}

int main(){
  while(cin >> X >> Y, X != 0 || Y != 0) {
    int sx = 21, sy;
    for(int i = 0; i < Y; i++){
      cin >> a[i];
      if(sx == 21)
	for(int j = 0; j < X; j++)
	  if(a[i][j] == '@') sy = i, sx = j;
    }
    cnt = 0;
    dfs(sy, sx);
    
    cout << cnt << endl;
  }
}