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
#include<bits/stdc++.h>
#define randome_generate srand(std::chrono::high_resolution_clock::now().time_since_epoch().count())
using namespace std;
char grid[1000][1000];
int n,m,k;
int dist[1000][1000];
queue<pair<int,int> >q;
int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>m>>k;
	memset(dist,63,sizeof(dist));
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cin>>grid[i][j];
			if(grid[i][j]=='S'){
				q.push(make_pair(i,j));
				dist[i][j]=0;
			}
		}
	}
	int ans=2e9,res;
	while(q.size()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		ans=min(ans,min(min(x-1,n-x),min(y-1,m-y)));
		for(int i=0;i<4;i++){
			int nx=x+dx[i],ny=y+dy[i];
			if(nx>=1&&nx<=n&&ny>=1&&ny<=m){
				if(grid[nx][ny]=='.'){
					if(dist[nx][ny]>dist[x][y]+1){
						dist[nx][ny]=dist[x][y]+1;
						if(dist[nx][ny]<=k)
							q.push(make_pair(nx,ny));
					}
				}
			}
		}
	}
	res=(ans+k-1)/k+1;
	cout<<res<<endl;
	return 0;
}
