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 H,W,K,Sx,Sy,dis[810][810],dep[810][810],Ans=1e9;
bool g[810][810],vis[810][810];
queue<pair<int,int> > q;

void BFS()
{
	while (!q.empty()) q.pop();
	vis[Sx][Sy]=1;
	q.push(make_pair(Sx,Sy));
	while (!q.empty())
	{
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		Ans=min(Ans,dis[x][y]+1);
		if (dep[x][y]==K) continue;
		if (g[x-1][y]&&!vis[x-1][y]) dep[x-1][y]=dep[x][y]+1,vis[x-1][y]=1,q.push(make_pair(x-1,y));
		if (g[x+1][y]&&!vis[x+1][y]) dep[x+1][y]=dep[x][y]+1,vis[x+1][y]=1,q.push(make_pair(x+1,y));
		if (g[x][y-1]&&!vis[x][y-1]) dep[x][y-1]=dep[x][y]+1,vis[x][y-1]=1,q.push(make_pair(x,y-1));
		if (g[x][y+1]&&!vis[x][y+1]) dep[x][y+1]=dep[x][y]+1,vis[x][y+1]=1,q.push(make_pair(x,y+1));
	}
}

int main()
{
#ifdef h10
	freopen("C.in","r",stdin);
	freopen("C.out","w",stdout);
#endif
	int i,j;
	char c;
	scanf("%d%d%d",&H,&W,&K);
	for (i=1;i<=H;i++)
	for (j=1;j<=W;j++)
	{
		c=getchar();
		while (c!='.'&&c!='#'&&c!='S') c=getchar();
		if (c=='#') continue;
		g[i][j]=1;
		dis[i][j]=min(min((i-1+K-1)/K,(j-1+K-1)/K),min((H-i+K-1)/K,(W-j+K-1)/K));
		if (c=='S') Sx=i,Sy=j;
	}
	BFS();
	printf("%d\n",Ans);
}