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<vector>
#include<string>
#include<cstdlib>
#define cinf(n,x,y) for(int i=0;i<(n);i++) cin >> x[i] >> y[i];
typedef long long int ll;
using namespace std;

int w,h;
string f[20];
int cnt;

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

void dfs(int x,int y){
	f[y][x]='#';
	cnt++;
	for(int i=0;i<4;i++){
		int nx=x+dx[i];
		int ny=y+dy[i];

		if(nx<0||nx>=w)continue;
		if(ny<0||ny>=h)continue;
		if(f[ny][nx]!='.')continue;

		dfs(nx,ny);
	}
}

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