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<algorithm>
#include<string>
#include<cstring>
using namespace std;
char mas[30][30];
int W,H,sum;
int dx[4] = {0,-1,0,1};
int dy[4] = {1,0,-1,0};
void dfs(int x, int y);
int main(){
  int sx,sy;
  while(1){
    sum = 0;
    cin >> W >> H;
    if(W == 0 && H == 0)break;
    for(int i = 0; i < H; i++){
      for(int j = 0; j < W; j++){
	cin >> mas[i][j];
	if(mas[i][j] == '@'){
	  sx = j;
	  sy = i;
	}
      }
    }
    dfs(sx,sy);
    cout << sum << endl;
  }
}
void dfs(int x, int y){
  sum++;
  mas[y][x] = '#';
  for(int i = 0; i < 4; i++){
    int nx = x + dx[i];
    int ny = y + dy[i];
    if(nx >= 0 && nx < W && ny >= 0 && ny < H && mas[ny][nx] != '#'){
      dfs(nx,ny);
    }
  }
}