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>
using namespace std;
#define inf 1e9
int n,m,k;
int cx[5] = {0,0,1,-1};
int cy[5] = {1,-1,0,0};
int len[805][805];
char s[805][805];
queue<pair<int,int> > q;
bool check(int x,int y)
{
    if(x<0 || y<0 || x>=n || y>=m || s[x][y]=='#') return 0;
    return 1;
}
int near(int x,int y)
{
    return min(min(x,n-1-x),min(y,m-1-y));
}
int main()
{
    int i,j,x,y,sx,sy;
    int ans;
    scanf("%d%d%d",&n,&m,&k);
    for(i=0;i<n;i++) scanf(" %s",s[i]);
    for(i=0;i<n;i++) for(j=0;j<m;j++) if(s[i][j]=='S') {sx = i; sy = j;}

    for(i=0;i<n;i++) for(j=0;j<m;j++) len[i][j] = inf;
    ans = inf;
    len[sx][sy] = 0;
    q.push({sx,sy});
    while(!q.empty())
    {
        x = q.front().first; y = q.front().second; q.pop();
        ans = min(ans,(int)ceil((double)near(x,y)*1.0/k));
        if(len[x][y]==k) continue;
        for(i=0;i<4;i++)
        {
            if(check(x+cx[i],y+cy[i]) && len[x+cx[i]][y+cy[i]] > len[x][y] + 1)
            {
                len[x+cx[i]][y+cy[i]] = len[x][y] + 1;
                q.push({x+cx[i],y+cy[i]});
            }
        }
    }
    printf("%d",ans+1);
}