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 N 805
char mp[N][N];
int n,m,k;
int ans=1e9;
int xx[4]={-1,0,1,0},yy[4]={0,-1,0,1};

struct msg{int x,y,step;};
queue<msg >q;
int main(){
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(mp[i][j]=='S')
                q.push((msg){i,j,0});
    while(!q.empty()){
        msg v=q.front();q.pop();
        int x=v.x,y=v.y;
        ans=min(ans,(min(min(x-1,n-x),min(y-1,m-y))+k-1)/k+1);
        for(int i=0;i<4;i++){
            int nx=x+xx[i],ny=y+yy[i];
            if(1<=nx&&nx<=n&&1<=ny&&ny<=m&&mp[nx][ny]=='.'&&v.step<k){
                mp[nx][ny]='x';
                q.push((msg){nx,ny,v.step+1});
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}