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>
#define N 805
using namespace std;
queue<int> P,Q,D;
int n,m,k,u,v,B[N][N],ans=(1<<30); char A[N][N];
void ins(int x,int y,int z){ if(z>k||B[x][y]) return ; P.push(x),Q.push(y),D.push(z),B[x][y]=1;}
void flood()
{
	int x,y,z;
	P.push(u),Q.push(v),D.push(0),B[u][v]=1;
	while(!P.empty()){
		x=P.front(),y=Q.front(),z=D.front();
		P.pop(),Q.pop(),D.pop();
		if(x>1&&!B[x-1][y]&&A[x-1][y]!='#') ins(x-1,y,z+1);
		if(y>1&&!B[x][y-1]&&A[x][y-1]!='#') ins(x,y-1,z+1);
		if(x<n&&!B[x+1][y]&&A[x+1][y]!='#') ins(x+1,y,z+1);
		if(y<m&&!B[x][y+1]&&A[x][y+1]!='#') ins(x,y+1,z+1);
	  }
}
int main()
{
	int i,j;
	scanf("%d %d %d\n",&n,&m,&k);
	for(i=1;i<=n;i++) scanf("%s",A[i]+1);
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
			if(A[i][j]=='S') u=i,v=j;
	flood();
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
			if(B[i][j])
				ans=min((min(min(j-1,m-j),min(i-1,n-i))+k-1)/k+1,ans);
	cout<<ans;
	return 0;
}