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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#define N 810

using namespace std;

char a[N][N];
int dis[N][N];
int n,m,k,bx,by,Min,Ans;

void dfs(int x,int y,int s){
  if(s>k) return ;
  if(x==1||x==n||y==1||y==m) { puts("1"); exit(0); }
  if(dis[x][y]<=s) return ;
  dis[x][y]=s;
  int now=min(min(x-1,n-x),min(y-1,m-y));
  if(now<Min) Min=now;
  if(a[x+1][y]!='#') dfs(x+1,y,s+1);
  if(a[x-1][y]!='#') dfs(x-1,y,s+1);
  if(a[x][y+1]!='#') dfs(x,y+1,s+1);
  if(a[x][y-1]!='#') dfs(x,y-1,s+1);
}

inline int Ceil(int x,int y){
  if(x%y) return x/y+1;
  else return x/y;
}

int main(){
  scanf("%d%d%d",&n,&m,&k);
  for(int i=1;i<=n;i++) scanf("%s",a[i]+1);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
      if(a[i][j]=='S') bx=i,by=j;
  memset(dis,0x7f,sizeof(dis));
  Min=min(min(bx-1,n-bx),min(by-1,m-by));
  dfs(bx,by,0);
  Ans=1+Ceil(Min,k);
  printf("%d\n",Ans);
  return 0;
}