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;
vector<pair<long,long>> neighbor={{0,1},{0,-1},{1,0},{-1,0}};
long H,W,K;
vector<string>A;
long escape(long y,long x){  
  return min(min((x-1)/K,(W-x-2)/K),min((y-1)/K,(H-y-2)/K))+1;
}
int main(){
  cin>>H>>W>>K;A=vector<string>(H);
  long sy,sx;
  for(long i=0;i<H;i++){
    cin>>A[i];
    if(A[i].find("S")!=string::npos){sy=i;sx=A[i].find("S");}
  }
  if(sy==0||sy==H-1||sx==0||sx==W-1){cout<<0;return 0;}
  vector<pair<long,long>>C(1,pair<long,long>(sy,sx));
  vector<vector<long>>D(H,vector<long>(W,-1));D[sy][sx]=0;
  queue<pair<long,long>>q;q.push(pair<long,long>(sy,sx));
  while(q.size()){
    pair<long,long>f=q.front();q.pop();
    for(auto n:neighbor){      
      long y=f.first+n.first,x=f.second+n.second;
      if(0<=x&&x<W&&0<=y&&y<H&&A[y][x]!='#'&&D[y][x]==-1){
        D[y][x]=D[f.first][f.second]+1;      
        if(D[y][x]<=K){
          if(y==0||y==H-1||x==0||x==W-1){cout<<1;return 0;}
          if(D[y][x]<K)q.push(pair<long,long>(y,x));    
          C.push_back(pair<long,long>(y,x));
        }
      }
    }
  }
  long ans=LONG_MAX;
  for(auto c:C)ans=min(ans,1+escape(c.first,c.second));
  cout<<ans;
}