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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
bool map[810][810];
int dis[810][810];
#define x first
#define y second
int main()
{
	int n=gi(),i,j,t,m=gi(),k=gi(),x,y,d,ans=1<<30;char c;
	queue< pair<int,int> >q;
	for (i=1;i<=n;i++)
		for (j=1;j<=m;j++) {
			while ((c=getchar())!='S'&&c!='#'&&c!='.');
			if (c=='S') x=i,y=j;
			map[i][j]=c!='#';
		}
	q.push(make_pair(x,y));dis[x][y]=1;
	const int dx[4]={1,-1,0,0};
	const int dy[4]={0,0,1,-1};
	while (!q.empty()) {
		x=q.front().x,y=q.front().y;q.pop();
		if (x==1||x==n||y==1||y==n) return puts("1"),0;
		d=min(min(x-1,n-x),min(y-1,m-y));
		ans=min(ans,(d+k-1)/k+1);
		if (dis[x][y]<=k)
			for (t=0;t<4;t++)
				if (map[x+dx[t]][y+dy[t]]&&!dis[x+dx[t]][y+dy[t]]) {
					dis[x+dx[t]][y+dy[t]]=dis[x][y]+1;
					q.push(make_pair(x+dx[t],y+dy[t]));
				}
	}
	printf("%d\n",ans);
	return 0;
}