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>
#include<unistd.h>
#define int long long
const int inf=8938103643641919514ll;
const int mod=1000000007ll;
const int dd[]={0,-1,0,1,0};
using namespace std;
int a,b,c,d;
int p[800][800],q[800][800];
int kyori(pair<int,int> x){
	int n=x.first;
	n=min(n,x.second);
	n=min(n,a-x.first-1);
	n=min(n,b-x.second-1);
	return n;
}
signed main(){
	int i,j;
	for(i=0;i<800;i++)
		for(j=0;j<800;j++)
			q[i][j]=inf;
	int u,v;
	cin>>a>>b>>c;
	for(i=0;i<a;i++){
		string s;
		cin>>s;
		for(j=0;j<b;j++){
			if(s[j]=='#')
				p[i][j]=1;
			if(s[j]=='S'){
				u=i;
				v=j;
			}
		}
	}
	queue<pair<int,int> > que;
	vector<pair<int,int> > ve;
	que.push({u,v});
	q[u][v]=0;
	while(!que.empty()){
		auto x=que.front();
		que.pop();
		int z=q[x.first][x.second];
		if(z>c)
			break;
		ve.push_back(x);
		for(i=0;i<4;i++){
			int uu=x.first+dd[i];
			int vv=x.second+dd[i+1];
			if(uu>=0&&uu<a&&vv>=0&&vv<b&&p[uu][vv]==0&&q[uu][vv]>z+1){
				q[uu][vv]=z+1;
				que.push({uu,vv});
			}
		}
	}
	int ans=inf;
	for(auto x:ve)
		ans=min(ans,1+(kyori(x)+c-1)/c);
	cout<<ans<<endl;
}