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>
int n,m,k,ans=0x3f3f3f3f;
char s[807][807];
int d[807][807],q[807*807][2],ql=0,qr=0;
void mins(int&a,int b){if(a>b)a=b;}
int min(int a,int b){return a<b?a:b;}
void push(int x,int y,int z){
	if(!d[x][y]&&s[x][y]=='.'&&z<=k+1){
		d[x][y]=z;
		q[++qr][0]=x;
		q[qr][1]=y;
	}
}
int main(){
	scanf("%d%d%d",&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')s[i][j]='.',push(i,j,1);
	}
	while(ql!=qr){
		int x=q[++ql][0],y=q[ql][1],z=d[x][y]+1;
		if(x>1)push(x-1,y,z);
		if(x<n)push(x+1,y,z);
		if(y>1)push(x,y-1,z);
		if(y<n)push(x,y+1,z);
		mins(ans,1+(min(min(x-1,n-x),min(y-1,m-y))+k-1)/k);
	}
	printf("%d\n",ans);
	return 0;
}