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;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int H,W,vis[1000][1000],K;
int mps[1000][1000],lds;
char sd[1000];
int sx,sy,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
inline void deal(int xx,int yy)
{
	lds=min(lds,xx-1);
	lds=min(lds,H-xx);
	lds=min(lds,yy-1);
	lds=min(lds,W-yy);
	
}
int main()
{
	H=read(); W=read(); K=read();
	lds=1000;
	for(int i=1;i<=H;i++)
	{
		scanf("%s",sd);
		for(int j=0;j<W;j++)
		{
			mps[i][j+1]=sd[j]=='#';
			if(sd[j]=='S') sx=i,sy=j+1;
		}
	}
	queue <int> q; while(!q.empty()) q.pop();
	//cerr << sx << " " << sy << endl;
	q.push(sx); q.push(sy); vis[sx][sy]=0;
	deal(sx,sy);
	while(!q.empty())
	{
		int lx=q.front(); q.pop();
		int ly=q.front(); q.pop();
		for(int i=0;i<4;i++)
		{
			int vx=lx+dir[i][0];
			int vy=ly+dir[i][1];
			if(vx<1||vy<1||vx>H||vy>W) continue;
			if(mps[vx][vy]) continue;
			if(!vis[vx][vy]&&vis[lx][ly]<K)
			{
				vis[vx][vy]=vis[lx][ly]+1;
				deal(vx,vy);
				q.push(vx); q.push(vy);
			}
		}
	}
	cout << 1+lds/K+(lds%K!=0) << endl;
	return 0;
}