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;
const int maxn=801;
int n,m,x,y,ind[maxn],k,b[maxn][maxn];
char s[maxn][maxn];
queue<int> qx,qy,qk;
void put(int x,int y,int k)
{
	if(b[x][y]||x<0||x>n||y<0||y>m||k<0||s[x][y]=='#')return;
	b[x][y]=1;qx.push(x);qy.push(y);qk.push(k);
}
void bfs(int xx,int yy,int kk)
{
	qx.push(xx),qy.push(yy);qk.push(kk),b[xx][yy]=1;
	while(!qx.empty())
	{
		int x=qx.front(),y=qy.front(),k=qk.front();
		qx.pop();qy.pop();qk.pop();
		put(x+1,y,k-1);put(x-1,y,k-1);put(x,y+1,k-1);put(x,y-1,k-1);
	}
}
signed main()
{
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)cin>>(s[i]+1);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)if(s[i][j]=='S')x=i,y=j;
	bfs(x,y,k);
	int ans=n+m;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)
	{
		int x=min(min(i,n-i+1),min(j,m-j+1));
		x--;
		if(b[i][j])
			ans=min(ans,(int)ceil((double)x/k+1));
	}
	cout<<ans;
}
/*
7 7 2
#######
#######
##.....
###S###
##.#.##
###.###
#######
一个网格图，从一个起点出发。
有些格子上锁。
每一轮你都可以不断往一个已解锁的四个方向的相邻格子走，
最多走k次走完后你可以选择至多k个未解锁的格子，将它们解锁。
求最少多少轮能走到一个边界格子。
选定路线之后肯定就不会干无聊的事情，那么选定路线就好了。
首先每一轮肯定是开满下一轮要走的k个格子，在有剩余的情况下也不需要继续向前开。
即使早早地开完了格子，该走几轮还是几轮。
也就是说，如果确定了第一轮的终点，那么后面的事情都就非常简单了。
因此只需要找到第一轮的最优起点。
那么从S开始搜一轮，对于所有可行点直接搞一下就好了。
*/