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 INF 1000000009
using namespace std;

int n,m,k,x,y;
char c[1000][1000];
bool vis[1000][1000];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int main()
{
	cin>>n>>m>>k;
	for(int i=0;i<n;i++){
	    scanf("%s",&c[i]);
	    for(int j=0;j<m;j++) if(c[i][j]=='S') x=i,y=j;
    }
    int ans=min(min(x,y),min(n-x-1,m-y-1));
    vis[x][y]=true;
    if(ans%k==0) ans=ans/k;
    else ans=ans/k+1;
    int minv=ans+1;
    queue<pair<int,pair<int,int> > > q;
    q.push(make_pair(0,make_pair(x,y)));
    while(!q.empty())
    {
    	pair<int,pair<int,int> > t=q.front();q.pop();
    	if(t.first>=k) continue;
		for(int w=0;w<4;w++){
			int nx=t.second.first+dx[w],ny=t.second.second+dy[w];
			if(nx<0||ny<0||nx>=n||ny>=m) continue;
			if(c[nx][ny]=='#') continue;
			if(vis[nx][ny]) continue;
			vis[nx][ny]=true;
			int res=min(min(nx,ny),min(n-nx-1,m-ny-1));
			if(res%k==0) res=res/k;
			else res=res/k+1;
			minv=min(minv,res+1);
			q.push(make_pair(t.first+1,make_pair(nx,ny)));
		}
	}
	cout<<minv<<endl;
	return 0;
}
/*
7 7 1
######.
#####..
####..#
###S.##
##..###
#..####
..#####

HAHA!
*/