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>
using namespace std;
const int N=801,X[4]={-1,0,0,1},Y[4]={0,-1,1,0};
bool v[N][N];
int n,m,k,p[N*N],q[N*N],s,t,P,Q,f[N][N],A=N;
string S[N];
int main()
{
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)
		cin>>S[i];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(S[i][j-1]=='S')
				v[i][j]=true,p[++t]=i,q[t]=j;
	while(s<t)
	{
		P=p[++s],Q=q[s];
		for(int i=0;i<4;i++)
			if(1<=P+X[i]&&P+X[i]<=n&&1<=Q+Y[i]&&Q+Y[i]<=m&&S[P+X[i]][Q+Y[i]-1]=='.'&&!v[P+X[i]][Q+Y[i]])
				f[P+X[i]][Q+Y[i]]=f[P][Q]+1,v[P+X[i]][Q+Y[i]]=true,p[++t]=P+X[i],q[t]=Q+Y[i];
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(v[i][j]&&f[i][j]<=k)
				A=min(A,(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);
	cout<<A+1;
	return 0;
}