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
#include<bits/stdc++.h>
#define N 810
#define p_b push_back
#define m_p make_pair
#define m1 first
#define m2 second
#define inf 1e9

using namespace std;

typedef pair<int,int> PII;
pair<PII,int> q[N*N];
bool vis[N][N];
const int tx[4]={1,0,-1,0},ty[4]={0,1,0,-1};
int l,r,n,m,k,ans=inf;
char mp[N][N];

int main(){
	int i,j,x,y,d;
	scanf("%d%d%d",&n,&m,&k);
	for(i=1;i<=n;i++){
		scanf("%s",mp[i]+1);
		for(j=1;j<=m;j++)
		if(mp[i][j]=='S') q[1]=m_p(m_p(i,j),0);
	}
	
	x=q[1].m1.m1; y=q[1].m1.m2;
	ans=min(ans,(n-1-x)/k+1);
	ans=min(ans,(x-2)/k+1);
	ans=min(ans,(m-1-y)/k+1);
	ans=min(ans,(y-2)/k+1);
	l=1; r=1;
	while(l<=r){
		if(q[l].m2==k) break;
		for(i=0;i<4;i++){
			x=q[l].m1.m1+tx[i]; y=q[l].m1.m2+ty[i]; d=q[l].m2+1;
			if(x<1||x>n||y<1||y>m||vis[x][y]||mp[x][y]=='#') continue;
			if(x==1||x==n||y==0||y==m){puts("1"); return 0;}
			vis[x][y]=1; q[++r]=m_p(m_p(x,y),d);
			ans=min(ans,(n-1-x)/k+1);
			ans=min(ans,(x-2)/k+1);
			ans=min(ans,(m-1-y)/k+1);
			ans=min(ans,(y-2)/k+1);
		}
		l++;
	}
	printf("%d\n",ans+1);
	return 0;
}