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;
typedef long long ll;
typedef double D;
typedef pair<int,int> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 1000000000
int h,w,k,d[1005][1005],sx,sy,dx[4]={0,1,0,-1},dy[4]={1,0,-1,0},ans=INF;
int main(void){
    scanf("%d%d%d",&h,&w,&k);
    for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			char c;
			scanf(" %c",&c);
			if(c=='S')sx=j,sy=i;
			if(c=='#')d[i][j]=INF;
			else d[i][j]=-1;
		}
	}
	queue<P>bfs;
	d[sy][sx]=0;
	bfs.push(P(sy,sx));
	while(!bfs.empty()){
		int x=bfs.front().S,y=bfs.front().F;
		bfs.pop();
		for(int i=0;i<4;i++){
			int nx=x+dx[i],ny=y+dy[i];
			if(nx>=0&&nx<w&&ny>=0&&ny<h&&d[ny][nx]==-1){
				d[ny][nx]=d[y][x]+1;
				bfs.push(P(ny,nx));
			}
		}
	}
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			if(d[i][j]!=-1&&d[i][j]<=k){
				int p=min(min(i,h-i-1),min(j,w-j-1));
				ans=min(ans,(p+k-1)/k+1);
			}
		}
	}
	printf("%d\n",ans);
}