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 <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int rd[4]={1,0,-1,0};
const int cd[4]={0,1,0,-1};

int h,w,k,sr,sc,d[805][805],res=inf;
char a[805][805];

inline bool inside(int r,int c){
    return r>=0 && r<h && c>=0 && c<w;
}

inline int casts(int r,int c){
    int x0=c,x1=w-c-1,y0=r,y1=h-r-1;
    return (min(min(x0,x1),min(y0,y1))+k-1)/k;
}

int main(){
    memset(d,inf,sizeof d);
    scanf("%d %d %d",&h,&w,&k);
    for(int i=0;i<h;i++){
        scanf("%s",a[i]);
        for(int j=0;j<w;j++)if(a[i][j]=='S')sr=i,sc=j;
    }
    queue<pair<int,int> > q;
    q.push(make_pair(sr,sc));
    d[sr][sc]=0;
    while(!q.empty()){
        pair<int,int> p=q.front();q.pop();
        res=min(res,casts(p.first,p.second)+1);
        if(d[p.first][p.second]==k)continue;
        for(int k=0;k<4;k++){
            int rr=p.first+rd[k],cc=p.second+cd[k];
            if(!inside(rr,cc) || a[rr][cc]=='#')continue;
            if(d[rr][cc]!=inf)continue;
            q.push(make_pair(rr,cc));
            d[rr][cc]=d[p.first][p.second]+1;
        }
    }
    printf("%d\n",res);
    return 0;
}