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 a[1005][1005];
int v[1005][1005];
int dis[1005][1005];
int sx,sy;
const int dx[]={-1,0,1,0},dy[]={0,-1,0,1};
struct Node{
	int x,y;
};
queue<Node>q;
inline int calc(int x,int y){
	int res=x-1;
	res=min(res,y-1);
	res=min(res,n-x);
	res=min(res,m-y);
	return res;
}
int main(){
	memset(dis,0x3f3f3f3f,sizeof(dis));
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			cin>>a[i][j];
			if(a[i][j]=='S'){
				q.push((Node){i,j});
				dis[i][j]=0;
			} 
		}
	while(!q.empty()){
		int x=q.front().x;
		int y=q.front().y;
		q.pop();
		v[x][y]=1;
		for(int i=0;i<4;i++){
			int xx=x+dx[i];
			int yy=y+dy[i];
			if(v[xx][yy]) continue;
			if(xx<1||xx>n||yy<1||yy>m) continue;
			if(a[xx][yy]=='#') continue;
			if(dis[xx][yy]>dis[x][y]+1){
				dis[xx][yy]=dis[x][y]+1;
				q.push((Node){xx,yy});
			}
		}
	}
	int ans=0x3f3f3f3f;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(dis[i][j]<=k){
				ans=min(ans,1+(calc(i,j)+k-1)/k);
			}
	cout<<ans<<endl;
	return 0;
}