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;
using ll=long long;
using P=pair<ll,ll>;
ll H,W,K;
ll dy[4]={-1,0,0,1};
ll dx[4]={0,1,-1,0};
char A[805][805];
ll B[805][805];
int main(){
  cin>>H>>W>>K;
  for(ll i=0;i<H;i++){
  string s;cin>>s;
    for(ll j=0;j<W;j++){
    A[i][j]=s[j];
    }
  }ll I,J;
  for(ll i=0;i<H;i++)
    for(ll j=0;j<W;j++)
      if(A[i][j]!='S')
        B[i][j]=-1;
  else{I=i;J=j;
    B[i][j]=0;
  }queue<P>que;P p11(I,J);que.push(p11);
  while(!que.empty()){
  P p=que.front();
    que.pop();
    ll y=p.first;ll x=p.second;
    for(ll i=0;i<4;i++)
      if(0<=dy[i]+y && dy[i]+y<H &&0<=dx[i]+x && dx[i]+x<W){
      ll Y=dy[i]+y;ll X=dx[i]+x;
        if(A[Y][X]=='.'&& B[Y][X]==-1){
          B[Y][X]=B[y][x]+1;
        P q(Y,X);que.push(q);
        }
      }
   }ll ans=10000000000;
  for(ll i=0;i<H;i++)
    for(ll j=0;j<W;j++)
      if(B[i][j]!=-1&& B[i][j]<=K){
      ll a=1+(min(min(i,H-1-i),min(j,W-1-j))+K-1)/K;
      if(a<ans)ans=a;
       }cout<<ans<<endl;
  return 0;
}