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 vis[801][801];

int H,W,K;
bool in(int x,int y){
	return x>=0&&x<H&&y>=0&&y<W;
}
string g[800];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>H>>W>>K;
	int sx,sy;
	for(int i=0;i<H;i++)
		cin>>g[i];
	for(int i=0;i<H;i++)
		for(int j=0;j<W;j++)
			if(g[i][j]=='S')
				sx=i,sy=j;
	queue<int> q;
	q.emplace(sx);
	q.emplace(sy);
	memset(vis,-1,sizeof vis);
	vis[sx][sy]=0;
	while(!q.empty()){
		int x=q.front();q.pop();
		int y=q.front();q.pop();
		if(vis[x][y]==K)	continue;
		for(int i=0;i<4;i++){
			int nx=x+dx[i];
			int ny=y+dy[i];
			if(in(nx,ny)&&vis[nx][ny]==-1&&g[nx][ny]=='.')
				vis[nx][ny]=vis[x][y]+1,q.emplace(nx),q.emplace(ny);
		}
	}
	int ans = 1e9;
	for(int i=0;i<H;i++)
		for(int j=0;j<W;j++)
			if(vis[i][j]!=-1)
				ans=min({ans,i,H-1-i,j,W-1-j});
	cout<<1+(ans+K-1)/K<<endl;
}