#include <bits/stdc++.h>
using namespace std;
char tile[20][21];
bool vis[20][20];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int W,H;
int dfs(int x,int y) {
    int sum=1;
    vis[y][x]=1;
    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||tile[ny][nx]=='#'||vis[ny][nx]) continue;
        sum+=dfs(nx,ny);
    }
    return sum;
}
int main() {
    while(scanf("%d %d",&W,&H),W) {
        for(int i=0;i<H;i++) scanf("%s",tile[i]);
        int sx,sy;
        for(int i=0;i<H;i++) {
            for(int j=0;j<W;j++) if(tile[i][j]=='@') {sx=j;sy=i;tile[i][j]='.';}
        }
        memset(vis,0,sizeof(vis));
        printf("%d\n",dfs(sx,sy));
    }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char tile[20][21];
bool vis[20][20];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int W,H;
int dfs(int x,int y) {
    int sum=1;
    vis[y][x]=1;
    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||tile[ny][nx]=='#'||vis[ny][nx]) continue;
        sum+=dfs(nx,ny);
    }
    return sum;
}
int main() {
    while(scanf("%d %d",&W,&H),W) {
        for(int i=0;i<H;i++) scanf("%s",tile[i]);
        int sx,sy;
        for(int i=0;i<H;i++) {
            for(int j=0;j<W;j++) if(tile[i][j]=='@') {sx=j;sy=i;tile[i][j]='.';}
        }
        memset(vis,0,sizeof(vis));
        printf("%d\n",dfs(sx,sy));
    }
}
```