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>

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)

using namespace std;
typedef pair<int,int> P;
const int INF=1e8;
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

const int MAX_W=800,MAX_H=800;

int H,W,K;
char A[MAX_H][MAX_W+1];

int dist[MAX_H][MAX_W];

int roundup(int x,int y){
	return (x+y-1)/y;
}

int main(){
	cin>>H>>W>>K;
	REP(i,H){
		cin>>A[i];
	}
	queue<P> que;
	REP(h,H){
		REP(w,W){
			dist[h][w]=INF;
			if (A[h][w]=='S'){
				que.push(P(h,w));
				dist[h][w]=0;
			}
		}
	}
	while(!que.empty()){
		P p=que.front();
		que.pop();
		int h=p.first;
		int w=p.second;
		REP(i,4){
			int h2=h+dy[i],w2=w+dx[i];
			if (0<=h2 && h2<H && 0<=w2 && w2<W && A[h2][w2]!='#' && dist[h][w]+1<dist[h2][w2]){
				dist[h2][w2]=dist[h][w]+1;
				que.push(P(h2,w2));
			}
		}
	}
	int ans=INF;
	REP(h,H){
		REP(w,W){
			if (dist[h][w]<=K){
				ans=min({ans,roundup(h,K),roundup(w,K),roundup(H-1-h,K),roundup(W-1-w,K)});
			}
		}
	}
	cout<<ans+1<<endl;
	return 0;
}