Problem: Takahashi is locked within a building.

This building consists of H×W rooms, arranged in H rows and W columns. We will denote the room at the i-th row and j-th column as (i,j). The state of this room is represented by a character A_{i,j}. If A_{i,j}= `#`, the room is locked and cannot be entered; if A_{i,j}= `.`, the room is not locked and can be freely entered. Takahashi is currently at the room where A_{i,j}= `S`, which can also be freely entered.

Each room in the 1-st row, 1-st column, H-th row or W-th column, has an exit. Each of the other rooms (i,j) is connected to four rooms: (i-1,j), (i+1,j), (i,j-1) and (i,j+1).

Takahashi will use his magic to get out of the building. In one cast, he can do the following:

* Move to an adjacent room at most K times, possibly zero. Here, locked rooms cannot be entered.
* Then, select and unlock at most K locked rooms, possibly zero. Those rooms will remain unlocked from then on.



His objective is to reach a room with an exit. Find the minimum necessary number of casts to do so.

It is guaranteed that Takahashi is initially at a room without an exit.

Constraints

* 3 ≤ H ≤ 800
* 3 ≤ W ≤ 800
* 1 ≤ K ≤ H×W
* Each A_{i,j} is `#` , `.` or `S`.
* There uniquely exists (i,j) such that A_{i,j}= `S`, and it satisfies 2 ≤ i ≤ H-1 and 2 ≤ j ≤ W-1.

Input

Input is given from Standard Input in the following format:


H W K
A_{1,1}A_{1,2}...A_{1,W}
:
A_{H,1}A_{H,2}...A_{H,W}


Output

Print the minimum necessary number of casts.

Examples

Input

3 3 3
#.#
#S.
###


Output

1


Input

3 3 3
.#
S.


Output

1


Input

3 3 3

S#


Output

2


Input

7 7 2


...##
S###
.#.##
.###


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

typedef std::pair<int,int>P;

int main(int argc, char *argv[])
{
  int H,W,K;
  char A[800][800];
  int d[800][800];
  P start;
  int distance=800;
  int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
  scanf("%d %d %d",&H,&W,&K);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      scanf(" %c",&A[i][j]);
      if(A[i][j]=='S'){
	d[i][j]=0;
	start.first=i;
	start.second=j;
	distance=std::min(std::min(i,H-i-1),std::min(j,W-j-1));
      }
    }
  }
  std::queue<P> que;
  que.push(start);
  while(!que.empty()){
    P now =que.front();que.pop();
    for (int i = 0; i < 4; i++) {
      int nx=now.first+dx[i];int ny=now.second+dy[i];
      if(nx>=0&&nx<H&&ny>=0&&ny<W&&A[nx][ny]=='.'){
	P next;
	next.first=nx;next.second=ny;
	que.push(next);
	d[nx][ny]=d[now.first][now.second]+1;
	A[nx][ny]='S';
	if(d[nx][ny]<=K){
	  distance=std::min(distance,std::min(std::min(nx,H-nx-1),std::min(ny,W-ny-1)));
	}
      }
    }
  }
  if (distance==0) {
    printf("1\n");
  }else{
    printf("%d\n",1+(distance+K-1)/K);
  }
  
  return 0;
}