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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,k,i,j,ans;
char mp[805][805];
int dis[805][805],sx,sy;
queue<int> qx,qy;
int main(){
	scanf("%d%d%d",&n,&m,&k);
	memset(dis,0x12,sizeof(dis));
	rep(i,n){
		scanf(" ");
		rep(j,m){
			scanf("%c",&mp[i][j]);
			if(mp[i][j]=='S'){
				sx=i;
				sy=j;
				if(i==1||i==n||j==1||j==m){
					puts("0");
					return 0;
				}
				dis[i][j]=0;
			}
		}
	}
	qx.push(sx);qy.push(sy);
	while(!qx.empty()){
		int x=qx.front(),y=qy.front();qx.pop();qy.pop();
		if(dis[x][y]>=k) continue;
		for(int nx=x-1;nx<=x+1;nx++)for(int ny=y-1;ny<=y+1;ny++){
			if(nx<1||nx>n||ny<1||ny>m) continue;
			if(abs(nx-x)+abs(ny-y)!=1) continue;
			if(mp[nx][ny]=='#') continue;
			if(dis[nx][ny]<=dis[x][y]+1) continue;
			dis[nx][ny]=dis[x][y]+1;
			qx.push(nx);qy.push(ny);
		}
	}
	ans=0x12121212;
	rep(i,n)rep(j,m) if(dis[i][j]<=k) ans=min(ans,1+(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);
	cout<<ans<<endl;
	return 0;
}