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 N 1116

int n,m,k,d[N][N];
char a[N][N];
queue<pair<int,int> > q;
int dx[]={-1,0,0,1},dy[]={0,-1,1,0};

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();k=read();
	memset(d,0x3f,sizeof(d));
	for (int i=1;i<=n;i++){
		scanf(" %s",a[i]+1);
		for (int j=1;j<=m;j++)
			if (a[i][j]=='S')
			d[i][j]=0,q.push(make_pair(i,j));
	}
	while (!q.empty()){
		int x=q.front().first,y=q.front().second;q.pop();
		for (int i=0;i<4;i++){
			int tx=x+dx[i],ty=y+dy[i];
			if (a[tx][ty]=='.'&&d[tx][ty]==0x3f3f3f3f)
			d[tx][ty]=d[x][y]+1,q.push(make_pair(tx,ty));
		}
	}
	int ans=0x3fffffff;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (d[i][j]<=k)
			ans=min(ans,1+(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);
	printf("%d\n",ans);
	return 0;
}