#include <stdio.h>
int A[4][2]={-1,0,0,1,1,0,0,-1},ans,N,M;char RB[25][25];void bfs(int x, int y) {RB[x][y]='#';ans++;for (int i=0;i<4;i++) {int nx=x+A[i][0],ny=y+A[i][1];if (nx>=0 && nx<N && ny>=0 && ny<M && RB[nx][ny]=='.') bfs(nx,ny);}}
int main() {int i,j;while (1) {scanf("%d %d",&M,&N);if (!N && !M) return 0;for (i=0;i<N;scanf("%s",&RB[i++]));ans=0;for (i=0;i<N;i++) for (j=0;j<M;j++) if (RB[i][j]=='@') bfs(i,j);printf("%d\n",ans);}}
### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <stdio.h>
int A[4][2]={-1,0,0,1,1,0,0,-1},ans,N,M;char RB[25][25];void bfs(int x, int y) {RB[x][y]='#';ans++;for (int i=0;i<4;i++) {int nx=x+A[i][0],ny=y+A[i][1];if (nx>=0 && nx<N && ny>=0 && ny<M && RB[nx][ny]=='.') bfs(nx,ny);}}
int main() {int i,j;while (1) {scanf("%d %d",&M,&N);if (!N && !M) return 0;for (i=0;i<N;scanf("%s",&RB[i++]));ans=0;for (i=0;i<N;i++) for (j=0;j<M;j++) if (RB[i][j]=='@') bfs(i,j);printf("%d\n",ans);}}
```