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>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
using P=pair<int,int>;
int inf=1e9;
queue<P> que;
vector<int>dx={0,0,1,-1};
vector<int>dy={1,-1,0,0};

int main(){
	int h,w,k;cin>>h>>w>>k;
	vector<string>a(h);
	for(auto&& u:a)cin>>u;
	vector<vector<int>>d(h,vector<int>(w,inf));
	rep(i,h)rep(j,w)if(a[i][j]=='S'){que.push(make_pair(i,j));d[i][j]=0;}
	while(!que.empty()){
		P z=que.front();que.pop();
		rep(i,4){
			int x,y;x=z.first+dx[i];y=z.second+dy[i];
			if(x<0||x>=h||y<0||y>=w)continue;
			if(d[x][y]==inf&&a[x][y]=='.' ){
				d[x][y]=d[z.first][z.second]+1;
				que.push(make_pair(x,y));
			}			

		}
	}	
	int mn=inf;
	rep(i,h)rep(j,w){if(d[i][j]<=k)mn=min({mn,i,j,h-1-i,w-1-j});}
	if (mn==0)cout<<1<<endl; 
	else  cout<<(mn-1)/k+2<<endl;
}