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>
using namespace std;
int n,m,k,d[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char mp[888][888];
bool vis[888][888];
struct node
{
	int x,y,num;
	node(int a=0,int b=0,int c=0) {x=a;y=b;num=c;}
};
queue<node>q;

void bfs(int x,int y)
{
	memset(vis,0,sizeof(vis));
	while(!q.empty()) q.pop();
	q.push(node(x,y,k));
	vis[x][y]=1;
	while(!q.empty())
	{
		node tp=q.front();q.pop();
		if(tp.num==0) continue;
		for(int i=0;i<4;++i)
		{
			int dx=tp.x+d[i][0],dy=tp.y+d[i][1];
			if(dx<0||dx>=n||dy<0||dy>=m) continue;
			if(mp[dx][dy]=='#') continue;
			if(vis[dx][dy]) continue;
			vis[dx][dy]=1;
			q.push(node(dx,dy,tp.num-1));
		}
	}
}
int cal(int L)
{
	if(L%k) return L/k+1;
	else return L/k;
}
int main()
{
    //freopen("3.txt","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<n;++i) scanf("%s",mp[i]);
	for(int i=0;i<n;++i)
		for(int j=0;j<m;++j) if(mp[i][j]=='S') bfs(i,j);
	int ans=2*n*m;
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<m;++j)
		{
			if(!vis[i][j]) continue;
			if(i==0||i==n-1||j==0||j==m-1) ans=min(ans,1);
			ans=min(ans,1+cal(i));
			ans=min(ans,1+cal(n-i-1));
			ans=min(ans,1+cal(j));
			ans=min(ans,1+cal(m-j-1));
		}
	}
	printf("%d\n",ans);
    return 0;
}