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 <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define Maxn 800
#define Inf 0x3f3f3f3f
int h,w,k,dist[Maxn+5][Maxn+5],sx,sy,d[5]={0,1,0,-1,0},ans;
char s[Maxn+5][Maxn+5];
queue<pair<int,int> > q;
int mn(int a,int b){
	return a<b?a:b;
}
int Get(int i,int j){
	int tmp=(i+k-2)/k;
	tmp=min(tmp,(j+k-2)/k);
	tmp=min(tmp,(h-i+k-1)/k);
	tmp=min(tmp,(w-j+k-1)/k);
	return tmp;
}
int main(){
	ans=Inf;
	scanf("%d%d%d",&h,&w,&k);
	for(int i=1;i<=h;i++){
		scanf("%s",s[i]+1);
	}
	for(int i=1;i<=h&&!sx;i++){
		for(int j=1;j<=w&&!sx;j++){
			if(s[i][j]=='S'){
				sx=i;
				sy=j;
			}
		}
	}
	memset(dist,0x3f,sizeof(dist));
	dist[sx][sy]=0;
	q.push(make_pair(sx,sy));
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		if(dist[x][y]==k){
			continue;
		}
		for(int i=1;i<=4;i++){
			int cx=x+d[i-1],cy=y+d[i];
			if(cx<1||cy<1||cx>h||cy>w||dist[x][y]+1>=dist[cx][cy]||s[cx][cy]=='#'){
				continue;
			}
			dist[cx][cy]=dist[x][y]+1;
			q.push(make_pair(cx,cy));
		}
	}
	for(int i=1;i<=h;i++){
		for(int j=1;j<=w;j++){
			if(dist[i][j]<Inf){
				ans=mn(ans,Get(i,j)+1);
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}