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;
const int MOD=1e9+7;
//const int MOD=998244353;
const int INF=1e9;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
int d[1000][1000];
//main
signed main(){
  int H,W,K;cin>>H>>W>>K;
  std::vector<string> s(H);
  for(int i=0;i<H;i++)cin>>s[i];
  int sx,sy;
  for(int i=0;i<H;i++)for(int j=0;j<W;j++)if(s[i][j]=='S')sy=i,sx=j;
  for(int i=0;i<H;i++)for(int j=0;j<W;j++)d[i][j]=LINF;
  int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};
  d[sy][sx]=0;
  typedef pair<int,int> P;
  queue<P> que;
  que.push(P(sy,sx));
  while(que.size()){
    P p=que.front();que.pop();
    for(int i=0;i<4;i++){
      int y=p.first+dy[i],x=p.second+dx[i];
      if(y<0||y>=H||x<0||x>=W)continue;
      if(s[y][x]=='#')continue;
      if(d[y][x]>d[p.first][p.second]+1){
        d[y][x]=d[p.first][p.second]+1;
        que.push(P(y,x));
      }
    }
  }
  int ans=INF;
  for(int i=0;i<H;i++)for(int j=0;j<W;j++)if(d[i][j]<=K)ans=min({ans,i,H-1-i,j,W-1-j});
  int res=ans/K+1;if(ans%K)res++;
  fin(res);
}