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 <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <stack>
#include <deque>
using namespace std;
#define ll long long
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define pii pair<pi,int>
#define sqr(x) ((x)*(x))
#define FOR for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int q[1010000][2],head,tail,d[810][810];
int n,m,k,sx,sy;
char p[810][810];
void ins(int x,int y,int cr){
	if(d[x][y]<=cr||p[x][y]!='.')return;
	d[x][y]=cr;
	tail++;
	q[tail][0]=x,q[tail][1]=y;}

int main()
{
	int i,j,x,y;
	cin>>n>>m>>k;
	for(i=1;i<=n;i++){
		scanf("%s",p[i]+1);
		for(j=1;j<=m;j++){
			if(p[i][j]=='S')

			sx=i,sy=j,p[i][j]='.';
			d[i][j]=1e9;}}
	ins(sx,sy,0);
	while (head<tail){
		head++;
		x=q[head][0],y=q[head][1];
		ins(x+1,y,d[x][y]+1);
		ins(x,y+1,d[x][y]+1);
		ins(x-1,y,d[x][y]+1);
		ins(x,y-1,d[x][y]+1);}
	int res=1e9;
	FOR{
		if(d[i][j]<=k)res=min(res,1+(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);}
	cout<<res<<endl;
    //system("pause");
    return 0;
}