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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;

const int N=1005;

struct node{
	int x,y;
	node(){}
	node(int x,int y):x(x),y(y){}
};
int n,m,d,sx,sy,x,y,ans;
int dx[4]={0,1,-1,0},dy[4]={1,0,0,-1},dep[N][N];
bool vis[N][N];
char s[N][N];
queue<node> q;

int main(){
	int i,j; node k;
	scanf("%d%d%d",&n,&m,&d);
	for (i=1;i<=n;i++){
		scanf("%s",s[i]+1);
		for (j=1;j<=m;j++) if (s[i][j]=='S') sx=i,sy=j;
	}
	q.push(node(sx,sy)); ans=n+m;
	vis[sx][sy]=1;
	while (!q.empty()){
		k=q.front(); q.pop();
		x=k.x; y=k.y;
		ans=min(ans,min(min(x-1,n-x),min(y-1,m-y)));
		if (dep[x][y]==d) continue;
		for (i=0;i<4;i++){
			x=k.x+dx[i]; y=k.y+dy[i];
			if (x<1||x>n||y<1||y>m) continue;
			if (vis[x][y]||s[x][y]!='.') continue;
			dep[x][y]=dep[k.x][k.y]+1;
			vis[x][y]=1;
			q.push(node(x,y));
		}
	}
	printf("%d\n",(ans+d-1)/d+1);
	return 0;
} 