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
#include<iostream>
#include<vector>
#include<string>

using namespace std;

int h,w;
string g[25];

int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

int dfs(int x, int y){
  g[y][x]='#';
  int ans=0;
  for(int i=0;i<4;i++){
    int nx=x+dx[i],ny=y+dy[i];
    if(0<=nx && nx<w && 0<=ny && ny < h && g[ny][nx]!='#'){
      ans+=dfs(nx,ny);
    }
  }
  return ans+1;
}

int main(void){

  while(cin >> w >> h, h|w){
    for(int i=0;i<h;i++)cin >> g[i];
    for(int i=0;i<h;i++){
      for(int j=0;j<w;j++){
	if(g[i][j]=='@')cout << dfs(j,i) << endl;
      }
    }
  }
  
  return 0;
}