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>
using namespace std;

int main()
{
  int h,w,K;
  scanf("%d%d%d",&h,&w,&K);
  static char M[810][810];
  for(int i=0;i<h;i++){
    scanf("%s",M[i]);
  }
  static int D[810][810];
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      D[i][j]=-1;
    }
  }
  queue<pair<int,int> > Q;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(M[i][j]=='S'){
	D[i][j]=0;
	Q.push(make_pair(i,j));
      }
    }
  }
  int ans;
  while(!Q.empty()){
    int i=Q.front().first,j=Q.front().second;
    Q.pop();
    if(i==0||i==h-1||j==0||j==w-1){
      ans=D[i][j];
      break;
    }
    int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
    for(int v=0;v<4;v++){
      int ni=i+dx[v],nj=j+dy[v];
      if(D[ni][nj]==-1&&(M[ni][nj]=='.'||D[i][j]>=K)){
	D[ni][nj]=D[i][j]+1;
	Q.push(make_pair(ni,nj));
      }
    }
    D[i][j]++;
    Q.push(make_pair(i,j));
  }
  printf("%d\n",(ans+K-1)/K);
  return 0;
}