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>
using namespace std;

void dfs(int x,int y);
int h,w;
char ta[22][22];
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int cou=0;

int main(){
  int a,b;
  while(1){
    cin >> w >> h;
    if(w==0 && h==0) break;
    cou=0;
    for(int i=0;i<h;i++){
      for(int j=0;j<w;j++){
	cin >> ta[i][j];
	if(ta[i][j]=='@'){
	  a=i,b=j;
	}
      }
    }
    dfs(a,b);
    cout << cou << endl;
  }
}

void dfs(int x,int y){
  ta[x][y]='#';
  cou++;
  for(int i=0;i<4;i++){
    int nx=x+dx[i];
    int ny=y+dy[i];
    if(nx>=0 && nx<h && ny>=0 && ny<w && ta[nx][ny]=='.'){
      dfs(nx,ny);
    }
  }
}