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>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int N = 810;
char s[N][N];
int x,y;
int f[N][N];
PII q[N*N];
int dx[4]={-1,1,0,0},dy[4]={0,0,1,-1};
bool in[N][N];
int main(){
	int H,W,K;
	cin>>H>>W>>K;
	for(int i =1 ;i<=H;i++) scanf("%s",s[i]+1);
	for(int i =1;i<=H;i++) for(int j  = 1;j<=W;j++) if(s[i][j]=='S'){
		x = i,y = j;
	}
	for(int i =0;i<N;i++) for(int j =0;j<N;j++) f[i][j]=1e9;
	f[x][y] = 0;
	int head = 0,tail = 0;
	q[0] = mp(x,y);
	int ans = 1e9;
	while(head<=tail){
		PII tmp = q[head++];
		//cout<<tmp.X<<' '<<tmp.Y<<endl;
		if(tmp.X==1 || tmp.X == H || tmp.Y ==1 || tmp.Y == W) ans = 1;
		int tmp2 = min(min(H-tmp.X,tmp.X-1),min(W-tmp.Y,tmp.Y-1));
		ans = min(ans,1+(tmp2+K-1)/K);
		for(int i =0;i<4;i++){
			x = tmp.X+dx[i];
			y = tmp.Y+dy[i];
			if(x>0 && x<=H && y>=0 && y<=W && f[tmp.X][tmp.Y]+1<=f[x][y] && s[x][y]=='.'){
				f[x][y] = f[tmp.X][tmp.Y]+1;
				if(f[x][y]<=K)
				if(!in[x][y]) {
					in[x][y] = 1;
					q[++tail] = mp(x,y);
				}
			}
		}
	}
	cout<<ans<<endl;
	
	return 0;
}