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
#include <bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N=805;

int n, m, k, ans=1e9;
char a[N][N];
int vis[N][N], dist[N][N], ways[N][N];
int dx[4]={-1, +1, 0, 0};
int dy[4]={0, 0, +1, -1};

void bfs(int x, int y, int n, int m)
{
	queue<pair<int, int> > q;
	q.push({x, y});
	ways[x][y]=1;
	vis[x][y]=1;
	dist[x][y]=0;
	while(!q.empty())
	{
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		if(a[x][y]=='#')
			continue;
		for(int dir=0;dir<4;dir++)
		{
			int nx=x+dx[dir];
			int ny=y+dy[dir];
			if(nx<1 || ny<1 || nx>n || ny>m)
				continue;
			if(!vis[nx][ny])
			{
				dist[nx][ny]=dist[x][y]+1;
				vis[nx][ny]=1;
				q.push({nx, ny});
			}
		}
	}
}

int get(int x, int y)
{
	return min({x-1, n-x, y-1, m-y});
}

int32_t main()
{
	IOS;
	int sx, sy;
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			cin>>a[i][j];
			if(a[i][j]=='S')
				sx=i, sy=j;
		}
	bfs(sx, sy, n, m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(vis[i][j] && dist[i][j]<=k && a[i][j]!='#')
				ans=min(ans, 1 + (get(i, j)+k-1)/k);
	cout<<ans;
	return 0;
}
