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<cstdio>
#include<algorithm>
using namespace std;
const int N=880;
int i,j,k,n,m,K,x,y,ans;
int c[N][N];
int bx[4]={1,-1,0,0};
int by[4]={0,0,1,-1};
int dx[N*N],dy[N*N];
char s[N][N];
int main() {
	scanf("%d%d%d",&n,&m,&K);
	for (i=1;i<=n;i++) {
		scanf("%s",s[i]+1);
		for (j=1;j<=m;j++) if (s[i][j]=='S') x=i,y=j;
	}
	i=0;j=1;
	dx[1]=x;dy[1]=y;
	c[x][y]=1;
	while (i<j) {
		i++;
		int dis=c[dx[i]][dy[i]];
		if (dis==K+1) break;
		for (k=0;k<4;k++) {
			int xx=dx[i]+bx[k];
			int yy=dy[i]+by[k];
			if (1<=xx && xx<=n && 1<=yy && yy<=m && !c[xx][yy] && s[xx][yy]!='#') {
				c[xx][yy]=dis+1;
				j++;
				dx[j]=xx;
				dy[j]=yy;
			}
		}
	}
	ans=n*m;
	for (i=1;i<=n;i++)
		for (j=1;j<=m;j++) if (c[i][j]) {
			int t1=min(i-1,n-i);
			int t2=min(j-1,m-j);
			ans=min(ans,(min(t1,t2)+K-1)/K);
		}
	printf("%d\n",ans+1);
}