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:
/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define all(v) v.begin(),v.end()
#define inf (int)(3e18)
#define P pair<int,int>
#define mod (int)(1e9+7)
template<class T> inline void chmin(T &a, T b) {
	a = min(a, b);
}
template<class T> inline void chmax(T &a, T b) {
	a = max(a, b);
}

int H,W,K;
char c[805][805];
int sx,sy;
int dis[805][805];
signed main(){
	cin>>H>>W>>K;
	rep(i,H)rep(j,W){
		cin>>c[i][j];
		dis[i][j]=inf;
		if(c[i][j]=='S'){
			sx=i,sy=j;
			dis[i][j]=0;
		}
	}
	queue<P>que;
	que.push({sx,sy});
	int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
	while(!que.empty()){
		P p=que.front();que.pop();
		if(dis[p.first][p.second]==K)break;
		rep(i,4){
			int nx=dx[i]+p.first,ny=dy[i]+p.second;
			if(nx<0||ny<0||nx>=H||ny>=W||c[nx][ny]=='#')continue;
			if(dis[nx][ny]==inf){
				dis[nx][ny]=dis[p.first][p.second]+1;
				que.push({nx,ny});
			}
		}
	}
	int ans=inf;
	rep(i,H)rep(j,W){
		if(dis[i][j]!=inf){
			int mi=min({i,H-i-1,j,W-j-1});
			chmin(ans,1+(mi+K-1)/K);
		}
	}
	cout<<ans<<endl;
}