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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <queue>

using namespace std;
typedef long long LL;
typedef pair<int,int> mp;
const int maxn = 805;
const int inf = 1<<30;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};

char s[maxn][maxn];
int g[maxn][maxn],h,w,k,cx,cy,ans=inf;
queue<mp> q;
bool inG(int x,int y) {
	return x>=1&&y>=1&&x<=h&&y<=w&&s[x][y]!='#';
}

int main()
{
	#ifdef Amberframe
		freopen("agc014c.in","r",stdin);
		freopen("agc014c.out","w",stdout);
	#endif
	scanf("%d %d %d",&h,&w,&k);
	for (int i=1;i<=h;i++) scanf("%s",s[i]+1);
	for (int i=1;i<=h;i++)
	for (int j=1;j<=w;j++) g[i][j]=inf;
	for (int i=1;i<=h;i++)
	for (int j=1;j<=w;j++)
		if (s[i][j]=='S') cx=i,cy=j;
	
	g[cx][cy]=0;q.push(mp(cx,cy));
	while (!q.empty()) {
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		if (g[x][y]>=k) break;
		for (int nx,ny,d=0;d<4;d++)
		if (inG(nx=x+dx[d],ny=y+dy[d])&&g[nx][ny]==inf)
			g[nx][ny]=g[x][y]+1,q.push(mp(nx,ny));
	}
	for (int i=1;i<=h;i++)
	for (int j=1;j<=w;j++)
		if (g[i][j]<=k) ans=min(ans,min(min(i-1,h-i),min(j-1,w-j)));
	printf("%d",ans?(ans-1)/k+2:1);
	return 0;
}