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;
const int N=805;
int n,m,k;
char s[N][N];
int d[N][N],qx[N*N],qy[N*N];
int dx[4]={ 0, 0,-1, 1};
int dy[4]={-1, 1, 0, 0};
bool check(int x,int y){
	return 1<=x&&x<=n&&1<=y&&y<=m&&s[x][y]!='#';
}
int main(){
	cin >> n >> m >> k;
	for (int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	memset(d,0,sizeof d);
	int sx,sy;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (s[i][j]=='S')
				sx=i,sy=j;
	int head=0,tail=0;
	tail++;
	qx[tail]=sx,qy[tail]=sy;
	d[sx][sy]=1;
	while (head<tail){
		head++;
		int x=qx[head],y=qy[head];
		if (d[x][y]==k+1)
			continue;
		for (int i=0;i<4;i++){
			int nx=x+dx[i],ny=y+dy[i];
			if (!check(nx,ny)||d[nx][ny])
				continue;
			d[nx][ny]=d[x][y]+1;
			tail++;
			qx[tail]=nx,qy[tail]=ny;
		}
	}
	int ans=23333;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (s[i][j]!='#'&&d[i][j]){
				if (i==1||j==1||i==n||j==m)
					ans=1;
				else {
					ans=min(ans,(i-1+k-1)/k+1);
					ans=min(ans,(j-1+k-1)/k+1);
					ans=min(ans,(n-i+k-1)/k+1);
					ans=min(ans,(m-j+k-1)/k+1);
				}
			}
	cout << ans;
	return 0;
}