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
//author: dxm
#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define INF 1e9
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
inline void Fail(){
	printf("0");
	exit(0);
}

const int maxn=805;
int n,m,k,mn=INF,sx,sy;
char grid[maxn][maxn];
int dis[maxn][maxn];
const int dx[]={1,0,-1,0};
const int dy[]={0,1,0,-1};

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

bool onboard(int x,int y){
	return x>=0&&x<n&&y>=0&&y<m;
} 

void bfs(){
	memset(dis,-1,sizeof(dis));
	dis[sx][sy]=0;
	queue<pii>q;
	q.push(mp(sx,sy));
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		if(dis[x][y]<=k)mn=min(mn,Dist(x,y));
		rep(i,4){
			int nx=x+dx[i],ny=y+dy[i];
			if(onboard(nx,ny)&&dis[nx][ny]==-1&&grid[nx][ny]=='.'){
				dis[nx][ny]=dis[x][y]+1;
				q.push(mp(nx,ny));
			}
		}
	}
}

int main(){
	scanf("%d%d%d",&n,&m,&k);
	rep(i,n){
		scanf("%s",grid[i]);
		rep(j,m)if(grid[i][j]=='S')sx=i,sy=j;
	}
	bfs();
	printf("%d",1+(mn+k-1)/k);
	return 0;
}

/*
Input:
7 7 1
#######
#######
##...##
#######
##.S.##
###.###
###.###
-----------------
Output:
*/