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;
int n,m,k;
char grid[1000][1000];
int steps[1000][1000];
queue<pair<int,int> > q;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
bool ok(int i,int j){
	return i>=0&&i<n&&j>=0&&j<m;
}
void BFS(int x,int y){
	memset(steps,0x3f,sizeof(steps));
	q.push(make_pair(x,y));
	steps[x][y]=0;
	while(!q.empty()){
		pair<int,int> cur=q.front();
		q.pop();
		for(int i=0;i<4;i++){
			int nx=cur.first+dx[i];
			int ny=cur.second+dy[i];
			if(ok(nx,ny)&&grid[nx][ny]!='#'&&steps[nx][ny]>=1e9){
				steps[nx][ny]=steps[cur.first][cur.second]+1;
				q.push(make_pair(nx,ny));
			}
		}
	}
}
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	cin>>n>>m>>k;
	pair<int,int> stt;
	for(int i=0;i<n;i++)
	{
		scanf("%s",grid[i]);
		for(int j=0;j<m;j++){
			if(grid[i][j]=='S')
				stt=make_pair(i,j);
		}
	}
	BFS(stt.first,stt.second);
	int res=1e9;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(grid[i][j]!='#'&&steps[i][j]<=k){
				int mmin=min(min(i,j),min(n-i-1,m-j-1));
				res=min(res,(mmin+k-1)/k+1);
			}
		}
	}
	cout<<res<<endl;
	return 0;
}