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<algorithm>
#include<queue>
using namespace std;
int ans,h,w,f,g,k;
bool canvis[810][810],vis[810][810];
queue<pair<int,int> > Q;
queue<int> lenQ;
void BFS(int y,int x,int len){
	if(vis[y][x]==true)return;
	vis[y][x]=true;
	int pnt=min(x,y);
	pnt=min(pnt,h-y-1);
	pnt=min(pnt,w-x-1);
	ans=min(ans,pnt);
	if(len==k)return;
	
	int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
	for(int i=0;i<4;i++){
		f=y+dy[i],g=x+dx[i];
		if(f>=0&&g>=0&&f<h&&g<w){
			if(canvis[f][g]&&!vis[f][g]){
				Q.push(make_pair(f,g));
				lenQ.push(len+1);
			}
		}
	}
}
int main(){
	cin>>h>>w>>k;
	ans=1145141919;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			char z;
			cin>>z;
			canvis[i][j]=true;
			if(z=='#')canvis[i][j]=false;
			if(z=='S')f=i,g=j;
			vis[i][j]=false;
		}
	}
	Q.push(make_pair(f,g));
	lenQ.push(0);
	while(!Q.empty()){
		BFS(Q.front().first,Q.front().second,lenQ.front());
		Q.pop(),lenQ.pop();
	}
	int outs=ans/k+1;
	if(ans%k>0)outs++;
	cout<<outs<<endl;
	return 0;
}