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 <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <numeric>
using namespace std;
typedef long long int ll;

char fi[850][850];
int d[850][850];

int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int h,w,k; cin >> h >> w >> k;
	int sx,sy;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			cin >> fi[i][j];
			d[i][j]=1e9;
			if(fi[i][j]=='S'){
				d[i][j]=0;
				sx=i,sy=j;
			}
		}
	}
	queue<pair<int,int>> q;
	q.push(make_pair(sx,sy));
	while(q.size()){
		auto p=q.front(); q.pop();
		int x=p.first,y=p.second;
		for(int i=0;i<4;i++){
			int nx=x+dx[i],ny=y+dy[i];
			if(0<=nx&&nx<h&&0<=ny&&ny<w&&fi[nx][ny]=='.'&&d[nx][ny]>d[x][y]+1){
				d[nx][ny]=d[x][y]+1;
				q.push(make_pair(nx,ny));
			}
		}
	}
	int res=1e9;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			if(d[i][j]<=k){
				int p=min({(i+k-1)/k,(h-1-i+k-1)/k,(j+k-1)/k,(w-1-j+k-1)/k});
				res=min(res,p+1);
			}
		}
	}
	printf("%d\n",res);
}