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 <iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
typedef pair<int,int> pint;
typedef long long ll;
const int inf   = 1<<30;
const int mod=1000000007;    
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};

int h,w,k,sx,sy,ans,dist[1000][1000];
string s[1000];

int mi(int a,int b,int c,int d,int e){
  return min(min(a,b),min(c,min(d,e)));
  }

void bfs(){
  queue<pint> q;
  q.push(pint(sx,sy));
  dist[sx][sy]=0;
  ans=mi(inf,sx,sy,h-1-sx,w-1-sy);
  while(!q.empty()){
    pint p=q.front();q.pop();
    int x=p.first,y=p.second;
    rep(i,4){
      int nx=x+dx[i],ny=y+dy[i];
      if(0<=nx&&nx<h&&0<=ny&&ny<w){
        if(s[nx][ny]=='.'&&dist[nx][ny]==inf){
          dist[nx][ny]=dist[x][y]+1;          ans=mi(ans,nx,ny,h-nx-1,w-ny-1);
          if(dist[nx][ny]<k)q.push(pint(nx,ny));
          }
        }
      }
    }
  }

          

int main(){
  cin>>h>>w>>k;
  rep(i,h){
    cin>>s[i];
    rep(j,w)if(s[i][j]=='S'){sx=i;sy=j;}
    }
  rep(i,h)rep(j,w)dist[i][j]=inf;
  bfs();
  if(ans)cout<<2+(ans-1)/k<<endl;
  else cout<<1<<endl;
  return 0;  
  }