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>
const int maxn=8e2+10;
using namespace std;
char a[maxn][maxn];
int stmp[maxn][maxn];
int direction[4][2]={1,0,-1,0,0,1,0,-1};
int n,m,k;
int ans=99999999;
bool check(int x,int y){
    if(x>=0 && x<n && y>=0 && y<m && stmp[x][y]==0 && a[x][y]=='.'){
       return 1;
    }
    return 0;
}
queue<pair<int,int> >po;
int main(){
    cin>>n>>m>>k;
    int sx,sy;
    getchar();
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            cin>>a[i][j];
            if(a[i][j]=='S'){
                sx=i;
                sy=j;
            }
        }
        getchar();
    }
    stmp[sx][sy]=1;
    po.push(make_pair(sx,sy));
    while(!po.empty()){
        int x=po.front().first;
        int y=po.front().second;
        ans=min(ans,min(min(x,n-x-1),min(y,m-y-1)));
        po.pop();
        for(int i=0;i<4;i++){
            int xx=x+direction[i][0];
            int yy=y+direction[i][1];
            if(check(xx,yy)&&stmp[x][y]<=k){
               stmp[xx][yy]=stmp[x][y]+1;
               po.push(make_pair(xx,yy));
            }
        }
    }
    int anss;
    anss=1+(ans+k-1)/k;
    cout<<anss<<endl;
    return 0;
}