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 REP(i,n,N) for(int i=(n);i<(int)N;i++)
#define p(s) cout<<(s)<<endl
#define ck(n,a,b) ((a)<=(n)&&(n)<(b))
#define F first
#define S second
typedef long long ll;
using namespace std;
const int inf=1e9;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
bool visit[810][810];
int main(){
	int H,W,K;
	cin>>H>>W>>K;
	char A[810][810];
	int sx,sy;
	REP(i,0,H) REP(j,0,W) {
		cin>>A[i][j];
		if(A[i][j]=='S'){
			sx=j;
			sy=i;
		}
	}
	queue<pair<int,pair<int,int>>> q;
	q.push({0,{sx,sy}});
	visit[sx][sy]=true;
	int mn=inf;
	while(!q.empty()){
		auto now=q.front();q.pop();
		REP(i,0,4){
			int nx=now.S.F+dx[i],ny=now.S.S+dy[i];
			int cost=now.F+1;
			if(!ck(ny,0,H)||!ck(nx,0,W)) continue;
			if(A[ny][nx]=='#'||visit[ny][nx]) continue;
			visit[ny][nx]=true;
			mn=min(mn,min(abs(nx-0),abs(W-1-nx)));
			mn=min(mn,min(abs(ny-0),abs(H-1-ny)));
			if(cost<K) q.push({cost,{nx,ny}});
		}
	}
	if(mn==inf) {
		mn=min(abs(sx-0),abs(W-1-sx));
		mn=min(mn,min(abs(sy-0),abs(H-1-sy)));
	}
	int ans=1;
	while(mn>0){
		ans++;
		mn-=K;
	}
	p(ans);
	return 0;
}