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<queue>
#include<algorithm>
using namespace std;
const int N=1005;
bool place[N][N];
bool used[N][N];
queue<int> q1;
int main(){
	int h,w,n;
	int i,j,k;
	int a,b,c;
	int p,q;
	char ca[N];
	cin>>h>>w>>n;
	for(i=0;i<h;i++){
		cin>>ca;
		for(j=0;j<w;j++){
			if(ca[j]!='#')place[i][j]=true;
			if(ca[j]=='S')p=i,q=j;
		}
	}
	q1.push(p*N+q);
	a=N;
	for(i=0;i<=n;i++){
		p=q1.size();
		for(j=0;j<p;j++){
			b=q1.front(),q1.pop();
			c=b%N,b/=N;
			if(used[b][c])continue;
			used[b][c]=true;
			if(!place[b][c])continue;
			a=min({a,b,c,h-b-1,w-c-1});
			if(b!=0)q1.push((b-1)*N+c);
			if(b!=h-1)q1.push((b+1)*N+c);
			if(c!=0)q1.push(b*N+c-1);
			if(c!=w-1)q1.push(b*N+c+1);
		}
	}
	if(a==0)a=1;
	else a=(a-1)/n+2;
	cout<<a<<endl;
	return 0;
}