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:
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define in insert
#define er erase
#define fr first
#define sc second
typedef long long ll;
const long long llinf=LLONG_MAX;
const int inf=INT_MAX;
const int nmax=1e5+5;
const int mod=1e9+7;
using namespace std;
int n,m,k,i,j,l,r,viz[801][801],ans=1e8;
pair<int,int>a[]={{1,0},{-1,0},{0,1},{0,-1}};
char c[801][801];
void solve(int i,int j)
{
	viz[i][j]=1;
	int d=min(min(i,j),min(n-i-1,m-j-1));
	ans=min(ans,(d+k-1)/k);
	for(int t=0;t<4;t++)if(i+a[t].fr>=0 && i+a[t].fr<n && j+a[t].sc>=0 && j+a[t].sc<m && !viz[i+a[t].fr][j+a[t].sc] && c[i+a[t].fr][j+a[t].sc]!='#' && abs(l-i-a[t].fr)+abs(r-j-a[t].sc)<=k)solve(i+a[t].fr,j+a[t].sc);
}
int main()
{
	//freopen("sol.in","r",stdin);
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n>>m>>k;
	for(i=0;i<n;i++)
	 for(j=0;j<m;j++)
	 {
	 	cin>>c[i][j];
	 	if(c[i][j]=='S')l=i,r=j;
	 }
	solve(l,r);
	cout<<ans+1<<endl;
	return 0;
}