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 <cstdio>
#include <cstring>
using namespace std;
int n,m,K,S,T;
pair<int,int> s[1000005];
char a[1005][1005];
int d[1005][1005];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
bool pan(int x,int y){
	if(x<1||y<1||x>n||y>m) return 0;
	return 1;
}
void bfs(int x,int y){
	if(x==1||y==1||x==n||y==m){
		puts("0");return;
	}
	memset(d,-1,sizeof(d));
	d[x][y]=0;
	s[0]=make_pair(x,y);
	int tou=0,wei=1,ans=1e9;
	while(tou<wei){
		x=s[tou].first;y=s[tou++].second;
		if(d[x][y]>K) break;
		ans=min(ans,min(min(x-1,y-1),min(n-x,m-y)));
		for(int i=0;i<4;i++) if(pan(x+dx[i],y+dy[i])&&d[x+dx[i]][y+dy[i]]==-1&&a[x+dx[i]][y+dy[i]]=='.'){
			d[x+dx[i]][y+dy[i]]=d[x][y]+1;
			s[wei++]=make_pair(x+dx[i],y+dy[i]);
		}
	}
	ans=ans==0?0:(ans-1)/K+1;
	printf("%d",ans+1);
}
int main(){
	scanf("%d%d%d",&n,&m,&K);
	for(int i=1;i<=n;i++) scanf("%s",a[i]+1);
	for(int i=1;i<=n;i++){
		for(int p=1;p<=m;p++) if(a[i][p]=='S') S=i,T=p;
	}
	bfs(S,T);
	return 0;
}