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>
typedef long long ll;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
ll n,m,k,i,j,d[1010][1010],has;
char a[1010][1010];
queue<pair<ll,ll> > q;
ll dx[4]={0,0,-1,1};
ll dy[4]={-1,1,0,0};
int main()
{
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m>>k;
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
		{
			cin>>a[i][j];
			if(a[i][j]=='S')
			{
				q.push(mp(i,j));
			}
			else
				d[i][j]=1e18;
		}
	has=1e18;
	while(!q.empty())
	{
		ll x=q.front().fi;
		ll y=q.front().se;
		if(d[x][y]<=k)
		{
		//	cout<<x<<" "<<y<<"_\n";
			has=min(has,min(min(x-1,y-1),min(n-x,m-y)));
		}
		q.pop();
		for(i=0;i<4;i++)
		{
			ll nx=x+dx[i];
			ll ny=y+dy[i];
			if(a[nx][ny]=='.'&&d[nx][ny]==1e18)
			{
		//		cout<<nx<<" "<<ny<<"\n";
				d[nx][ny]=d[x][y]+1;
				q.push(mp(nx,ny));
			}
		}
	}
	//cout<<has<<"\n";
	ll tam=0;
	if((has%k)!=0)
		tam=1;
	has=(has/k)+tam;
	has++;
	cout<<has<<"\n";
}