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
#include "bits/stdc++.h"
using namespace std;
#define int long long
int mod=1e9+7;
 
signed main(){
	int h,w,k;
	cin>>h>>w>>k;
	vector<string> a(h);
	for(int i=0;i<h;i++)cin>>a[i];
	
	queue<int> qx;
	queue<int> qy;
	vector<vector<int> > used(h,vector<int>(w,-1));
	
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			if(a[i][j]=='S'){
				qx.push(j);
				qy.push(i);
				used[i][j]=0;
			}
		}
	}
	
	int px[]={1,0,-1,0};
	int py[]={0,-1,0,1};
	
	int ans=1e9;
	
	
	while(!qx.empty()){
		int x=qx.front();
		int y=qy.front();
		qx.pop();
		qy.pop();
		if(used[y][x]<=k)ans=min(ans,min(x,min(y,min(w-x-1,h-y-1))));
		for(int i=0;i<4;i++){
			int nx=x+px[i];
			int ny=y+py[i];
			if(nx>=0&&ny>=0&&nx<w&&ny<h&&a[ny][nx]=='.'&&used[ny][nx]==-1){
				used[ny][nx]=used[y][x]+1;
				qx.push(nx);
				qy.push(ny);
			}
		}
	}
	
	int tmp=1;
	if(ans%k!=0)tmp++;
	tmp+=ans/k;
	cout<<tmp<<endl;
}