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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#define MN 800
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
char st[MN+5][MN+5];
int n,m,K,d[MN+5][MN+5],top=0,vis[MN+5][MN+5];
pair<int,int> q[MN*MN+5];
const int dis[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int main()
{
    n=read();m=read();K=read();
    for(int i=1;i<=n;++i) scanf("%s",st[i]+1);
    for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) if(st[i][j]=='S') q[++top]=make_pair(i,j),vis[i][j]=1;
    for(int i=1;i<=top;++i)
    {
        int x=q[i].first,y=q[i].second;
        for(int k=0;k<4;++k)
        {
            int nx=x+dis[k][0],ny=y+dis[k][1];
            if(nx<1||ny<1||nx>n||ny>m||vis[nx][ny]||st[nx][ny]=='#') continue;
            vis[nx][ny]=1;q[++top]=make_pair(nx,ny);d[nx][ny]=d[x][y]+1;
        }
    }
    int ans=1e9;
    for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) if(vis[i][j]) ans=min(ans,max(1,(d[i][j]+K-1)/K)+(min(min(i-1,n-i),min(j-1,m-j))+K-1)/K);
    cout<<ans;
    return 0;
}