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 <fstream>
#include <queue>
using namespace std;
const int nmax=805;
int d1[]={-1,0,1,0};
int d2[]={0,-1,0,1};
pair<int,int> q[nmax*nmax];
string s[nmax];
int d[nmax][nmax];
int n,m,k,i,j,l1,c1,p,u,li,ci,lf,cf,ans;
int main()
{
    cin>>n>>m>>k;
    for(i=1;i<=n;i++)
    {
        cin>>s[i];
        s[i]=" "+s[i];
        for(j=1;j<=m;j++)
            if(s[i][j]=='S')
        {
            l1=i;c1=j;
        }
        else
            if(s[i][j]=='#')
               d[i][j]=-1;
    }
    q[u=1]={l1,c1};
    d[l1][c1]=1;
    for(p=1;p<=u;p++)
    {
        li=q[p].first;ci=q[p].second;
        for(i=0;i<4;i++)
        {
            lf=li+d1[i];cf=ci+d2[i];
            if(lf>=1&&lf<=n&&cf>=1&&cf<=m&&(!d[lf][cf]))
            {
                q[++u]={lf,cf};
                d[lf][cf]=d[li][ci]+1;
            }
        }
    }
    ans=n*m+1;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
          if(d[i][j]>=1&&d[i][j]<=k+1)
          {
              ans=min(ans,(n-i+k-1)/k+1);
              ans=min(ans,(i-1+k-1)/k+1);
              ans=min(ans,(m-j+k-1)/k+1);
              ans=min(ans,(j-1+k-1)/k+1);
          }
    cout<<ans;
    return 0;
}