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>

#define F first
#define S second
#define pb push_back
#define ll long long
#define int long long

using namespace std ;

const int MAXN = 888 , MOD = 1e9 + 7 ; 

char ma[MAXN][MAXN] ; 

int d[MAXN][MAXN] ; 

int dx[] = {1,-1,0,0} ; 
int dy[] = {0,0,1,-1} ; 

int32_t main()
{
	ios::sync_with_stdio(0) ; cin.tie(0) ; 
 	int n , k ,  m ; 
	cin >> n >> m >> k  ; 
	int ans = 1e18 , x , y ; 
	for(int i = 0 ; i < n ; i ++ ) for(int j = 0 ; j < m ; j ++ ) 
	{
		cin >> ma[i][j] ;
		if(ma[i][j]=='S') x = i , y = j ; 
	}
	queue<pair<int,int>> q ; 
	q.push({x,y}) ;
	memset(d,63,sizeof d) ; 
	d[x][y] = 0 ; 
	while(q.size())
	{
		int x = q.front().F , y = q.front().S ; 
		q.pop() ; 
		if(d[x][y]>k) break ;
		int i = x , j = y ;
		ans = min(ans,(i+k-1)/k) ; 
		ans = min(ans,(j+k-1)/k) ; 
		ans = min(ans,(n-i-1+k-1)/k) ; 
		ans = min(ans,(m-j-1+k-1)/k) ; 	
		for(int i = 0 ; i < 4;  i ++ ) 
		{
			int X = x + dx[i] , Y = y + dy[i] ; 
			if(X<0||X>=n||Y<0||Y>=m||d[X][Y]<=d[x][y]+1||ma[X][Y]=='#') continue ; 
			d[X][Y] = d[x][y] + 1 ; 
			q.push({X,Y}) ; 
		}
	}
	cout << ans + 1<< endl ; 

}