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 <cstdio>
#include <cstring>
#include <algorithm>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

const int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};
int n,m,k,d[810][810],q[640100][2],head,tail,ans;
char a[810][810];

void bfs(int x,int y)
{
    memset(d,255,sizeof(d)),d[x][y]=0;
    q[0][0]=x,q[0][1]=y;
    for (; head<=tail; ++head)
    {
        x=q[head][0],y=q[head][1];
        if (d[x][y]<=k)
            ans=min(ans,(min(min(x-1,y-1),min(n-x,m-y))+k-1)/k+1);
        repu(i,0,3)
            if (x+fx[i] && x+fx[i]<=n && y+fy[i] && y+fy[i]<=m
            && a[x+fx[i]][y+fy[i]]!='#' && d[x+fx[i]][y+fy[i]]==-1)
            {
                d[x+fx[i]][y+fy[i]]=d[x][y]+1;
                q[++tail][0]=x+fx[i],q[tail][1]=y+fy[i];
            }
    }
}

int main()
{
    scanf("%d%d%d",&n,&m,&k);
    repu(i,1,n)
        scanf("%s",a[i]+1);
    repu(i,1,n)
        repu(j,1,m)
            if (a[i][j]=='S')
                ans=1<<30,bfs(i,j);
    printf("%d\n",ans);
    return 0;
}