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;
typedef long long ll;
#define H 800

int vis[H][H];
char a[H][H];
int h,w,k;
int u,d,l,r;

bool check(int x,int y,int k){
	if(x>=0 && x<h && y>=0 && k>=1 && y<w && a[x][y]!='#' && !vis[x][y])return true;
	return false;
}

int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

void dfs(int x,int y,int k){
	queue<pair<int,int>> que;
	queue<int> kque;
	que.push({x,y});
	vis[x][y]=1;
	kque.push(k);
	while(que.size()){
		x=que.front().first;
		y=que.front().second;
		k=kque.front();
		que.pop();
		kque.pop();
		u=min(x,u);
		d=max(x,d);
		l=min(l,y);
		r=max(r,y);
		for (int i = 0; i < 4; i++) {
			if(check(x+dx[i],y+dy[i],k)){
				que.push({x+dx[i],y+dy[i]});
				vis[x+dx[i]][y+dy[i]]=1;
				kque.push(k-1);
			}
		}
	}
}

int main(void)
{
	cin>>h>>w>>k;
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			cin>>a[i][j];
			if(a[i][j]=='S'){
				u=d=i;
				l=r=j;
			}
		}
	}
	dfs(u,l,k);
	int dmin=min(u-0,min(h-1-d,min(l-0,w-1-r)));
	cout<<ceil(float(dmin)/k)+1<<endl;
	return 0;
}