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 x first
#define y second
#define N 880
using namespace std;
typedef pair<int,int> pii;
char dt[N][N];
int dis[N][N];
int n,m,k;
inline bool pc(int x,int y){
	if(x==n||y==m||x<0||y<0||dt[x][y]=='#'||dis[x][y]!=-1) return 0;
	return 1;
}
queue<pii> q;
int fx[4]={1,-1,0,0},fy[4]={0,0,1,-1};
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<n;i++) scanf("%s",dt[i]);
	memset(dis,-1,sizeof(dis));
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++)
			if(dt[i][j]=='S') q.push(pii(i,j)),dis[i][j]=0;;
	}
	int ans=1008610086;
	while(!q.empty()){
		pii st=q.front();q.pop();
		if(dis[st.x][st.y]>k) continue;
		ans=min(min(st.x,st.y),min(min(n-st.x,m-st.y)-1,ans));
		for(int i=0;i<4;i++){
			pii to=pii(st.x+fx[i],st.y+fy[i]);
			if(pc(to.x,to.y)){
				q.push(to);
				dis[to.x][to.y]=dis[st.x][st.y]+1;
			}
		}
	}
	printf("%d\n",(ans+k-1)/k+1);
	return 0;
}