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;
#define nn 808
int n,m,k,dist[nn][nn];
char locked[nn][nn];

int ax[]={0,1,-1,0},ay[]={-1,0,0,1};
void bfs(){
	queue<int> qx,qy;
	memset(dist,0x3f,sizeof dist);
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(locked[i][j]=='S') 
					qx.push(i),qy.push(j),dist[i][j]=0;
	
	while(qx.size()){
		int x=qx.front(),y=qy.front();qx.pop(),qy.pop();
		for(int i=0;i<4;i++){
			int xx=ax[i]+x,yy=ay[i]+y;
			
			if(locked[xx][yy]=='.' and dist[xx][yy]>dist[x][y]+1){
				dist[xx][yy]=dist[x][y]+1;
				qx.push(xx),qy.push(yy);
			}
		}
	}
}

signed main(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++) scanf("%s",locked[i]+1);
	bfs();int ans=6489648;
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(dist[i][j]<=k)	
		ans=min(ans,1+(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);

	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(i==1 or i==n or j==1 or j==m)
		ans=min(ans,(dist[i][j]+k-1)/k);
	cout<<ans;
	return 0;
}