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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=810,M=N*N;
bool w[N][N];
char s[N][N];
int a[N],b[N],c[N],m1[4]={1,0,-1,0},m2[4]={0,1,0,-1};
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main() {
    int n,m,k,ans=1<<30,x,y;
    cin>>n>>m>>k;
    for(int i=1;i<=n;i++) {
	scanf("%s",s[i]+1);
	for(int j=1;j<=m;j++)
	    if(s[i][j]=='S') x=i,y=j;
    }
    int t=0,sum=1;
    a[1]=x,b[1]=y,w[x][y]=1;
    while(t<sum) {
	x=a[++t],y=b[t];
	ans=min(ans,(min(min(x-1,n-x),min(y-1,m-y))+k-1)/k+1);
	if(c[t]==k) continue;
	for(int i=0;i<4;i++) {
	    int X=x+m1[i],Y=y+m2[i];
	    if(!w[X][Y]&&s[X][Y]=='.') {
		c[++sum]=c[t]+1,w[X][Y]=1;
		a[sum]=X,b[sum]=Y;
	    }
	}
    }
    cout<<ans;
    return 0;
}