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
#include<bits/stdc++.h>
using namespace std;

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

int H,W,K;
char t[1000][1000];
int d[1000][1000];

int si,sj;

void bfs(){

  int ans=1e9;
  
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      d[i][j]=1e9;
  
  queue<int> qy,qx;
  qy.push(si);
  qx.push(sj);
  d[si][sj]=0;
  
  while(!qy.empty()){
    int y=qy.front();
    qy.pop();
    int x=qx.front();
    qx.pop();

    if(d[y][x]>K)continue;
    
    int f=min(y,x);
    f=min(f,H-1-y);
    f=min(f,W-1-x);

    if(f==0)ans=1;
    
    ans=min(ans,1 + (f+K-1)/K);
    for(int i=0;i<4;i++){
      int ny=y+dy[i];
      int nx=x+dx[i];
      if(ny<0||nx<0||ny>=H||nx>=W)continue;
      if(t[ny][nx]=='#')continue;
      int ncost=d[y][x]+1;
      if(d[ny][nx]>ncost){
        d[ny][nx]=ncost;
        qy.push(ny);
        qx.push(nx);
      }
    }
  }
  /*
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      cout<<d[i][j]<<' ';
    }
    cout<<endl;
  }
  */
  cout<<ans<<endl;
}

int main(){
  cin>>H>>W>>K;
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      cin>>t[i][j];
      if(t[i][j]=='S'){
        si=i;
        sj=j;
      }
    }
  }
  bfs();
  return 0;
}
