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>
using namespace std;
struct data{
	int x,y;
	data(){}
	data(int x,int y):x(x),y(y){}
}q[1001000];
int dist[810][810],k,dx[4]={0,0,1,-1},dy[4]={1,-1,0,0},w,h,l,r;
char s[810][810];
int get(int x){
	if(x%k==0)return x/k;
	else return x/k+1;
}
int main(){
	scanf("%d%d%d",&w,&h,&k);
	for(int i=1;i<=w;++i)
		scanf("%s",s[i]+1);
	for(int i=1;i<=w;++i)
		for(int j=1;j<=h;++j)
			if(s[i][j]=='S')q[r++]=data(i,j),dist[i][j]=1;
	while(l<r){
		data d=q[l++];
		if(dist[d.x][d.y]==k+1)continue;
		for(int i=0;i<4;++i){
			int nx=d.x+dx[i],ny=d.y+dy[i];
			if(nx<1||ny<1||nx>w||ny>h)continue;
			if(dist[nx][ny]||s[nx][ny]!='.')continue;
			dist[nx][ny]=dist[d.x][d.y]+1;
			q[r++]=data(nx,ny);
		}
	}
	int ans=1<<30;
	for(int i=1;i<=w;++i)
		for(int j=1;j<=h;++j)if(dist[i][j])
			ans=min(ans,get(min(min(i-1,j-1),min(w-i,h-j)))+1);
	printf("%d",ans);
}