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<bits/stdc++.h>
#define fi first
#define se second

using namespace std;
typedef pair<int,int>pi;

const int N=805,M=640005;

char s[N][N];
int n,m,K,ans,dis[N][N],ql,qr;
pi qu[M];

void bfs(int ux,int uy){
	memset(dis,-1,sizeof(dis));
	qu[ql=qr=1]={ux,uy};dis[ux][uy]=0;
	while(ql<=qr&&dis[qu[ql].fi][qu[ql].se]<=K){
		int x=qu[ql].fi,y=qu[ql].se;
		for(int dx:{-1,0,1})for(int dy:{dx?0:-1,dx?0:1})
			if(s[x+dx][y+dy]=='.'&&!~dis[x+dx][y+dy])
				dis[x+dx][y+dy]=dis[x][y]+1,qu[++qr]={x+dx,y+dy};
		ql++;ans=min(ans,min(min(x-1,n-x),min(y-1,m-y)));
	}
}

int main(){
	scanf("%d%d%d",&n,&m,&K);ans=min(n,m);
	for(int i=1;i<=n;i++)scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(s[i][j]=='S'){bfs(i,j);break;}
	printf("%d\n",1+(ans+K-1)/K);
	return 0;
}