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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const int N=1e3+5;
char s[N][N];
int n,m,k;
int sx,sy;
int z[N][N];
int ans=1<<30;
typedef pair<int,int>pi;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
queue<pi>q;
bool vis[N][N];
int deep[N][N];
void bfs(){
    q.push(pi(sx,sy));
    vis[sx][sy]=1;
    while(!q.empty()){
        int x=q.front().first,y=q.front().second;
        q.pop();
        ans=min(ans,z[x][y]);
        if(deep[x][y]==k)
            continue;
        for(int i=0;i<4;i++){
            int xx=x+dir[i][0],yy=y+dir[i][1];
            if(xx<=0||yy<=0||xx>n||yy>m||vis[xx][yy]||s[xx][yy]=='#')
                continue;
            vis[xx][yy]=1;
            deep[xx][yy]=deep[x][y]+1;
            q.push(pi(xx,yy));
        }
    }
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++){
        scanf("%s",s[i]+1);
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            if(s[i][j]=='S')
                sx=i,sy=j;
            z[i][j]=min(min(i-1,j-1),min(n-i,m-j));
            z[i][j]=(z[i][j]+k-1)/k;
        }
    bfs();
    printf("%d\n",ans+1);
}