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;
int dist[801][801];
string f[801];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
template<class T> bool uin(T &a,T b){return a>b?(a=b,true):false;}
template<class T> bool uax(T &a,T b){return a<b?(a=b,true):false;}
int nceil(int a,int b){
	return (a+b-1)/b;
}
int main(){
	int n,m,k;
	cin>>n>>m>>k;
	int sx=-1,sy=-1;
	for(int i=0;i<n;i++){
		cin>>f[i];
		for(int j=0;j<m;j++){
			if(f[i][j]=='S'){
				sx=i;
				sy=j;
			}
		}
	}
	vector<pair<int,int> > q;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			dist[i][j]=1e9;
		}
	}
	dist[sx][sy]=0;
	q.push_back(make_pair(sx,sy));
	for(int cur=0;cur<q.size();cur++){
		int x=q[cur].first,y=q[cur].second;
		int ncur=dist[x][y];
		for(int d=0;d<4;d++){
			int xx=x+dx[d],yy=y+dy[d];
			if(min(xx,yy)<0||min(n-xx,m-yy)<=0||dist[xx][yy]<1e8||f[xx][yy]=='#')continue;
			dist[xx][yy]=ncur+1;
			q.push_back(make_pair(xx,yy));
		}
	}
	int ans=1e9;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(dist[i][j]>k)continue;
			int res=1e9;
			uin(res,nceil(i,k));
			uin(res,nceil(j,k));
			uin(res,nceil(n-i-1,k));
			uin(res,nceil(m-j-1,k));
			uin(ans,res+1);
		}
	}
	cout<<ans<<endl;
	return 0;
}