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 <iostream> 
#include <string> 
#include <vector> 
#include <cmath> 
#include <algorithm> 
#include <cstdlib> 
#include <ctime> 
#include <cstdio> 
#include <functional> 
#include <set> 
#include <sstream> 
#include <cctype>
#include <stack>
#include <queue>
#include <cstring>
#include <map>
#include <list>
#include <cassert>
 
using namespace std; 


int dy[]={-1,0,1,0};
int dx[]={0,1,0,-1};
int h,w,K;
int sy,sx;
int res=1e8;
bool vis[800][800];
void bfs(vector<string> &a,int y,int x){

	
	vis[y][x]=true;
	int d=min({y,x,h-y-1,w-x-1});
	res=min(res,(d+K-1)/K);

	for(int i=0;i<4;i++){
		int ny=y+dy[i],nx=x+dx[i];
		if(0<=ny && ny<h && 0<=nx && nx<w && !vis[ny][nx] && a[ny][nx]!='#' && abs(sy-ny)+abs(sx-nx)<=K ){
			bfs(a,ny,nx);
		}
	}


}

int main(){

	
	cin>>h>>w>>K;

	vector<string> a(h);
	
	for(int i=0;i<h;i++){
		cin>>a[i];
		for(int j=0;j<w;j++){
			if(a[i][j]=='S') sy=i,sx=j;
		}
	}
	
	bfs(a,sy,sx);
	cout<<res+1<<endl;

	return 0;
}