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 <iostream>
#include <queue>
#include <string>
#include <vector>
#include <algorithm>
#include <tuple>
using namespace std;

int main()
{
    int n, m, k; cin>>n>>m>>k;

    vector<string> a(n);
    vector<vector<int>> v(n, vector<int>(m, -1));
    queue<pair<int, int>> q;

    for(int i=0; i<n; ++i) {
        cin>>a[i];
        for(int j=0; j<m; ++j) {
            if (a[i][j]=='S') {
                q.emplace(i, j);
                v[i][j]=0;
                break;
            }
        }
    }

    int dr[4]={-1, 0, 1, 0};
    int dc[4]={0, -1, 0, 1};
    int res=1<<30;
    while (q.size()) {
        int r, c;
        tie(r, c)=q.front();
        q.pop();
        //res=min({res, r, c, n-1-r, m-1-c});
        res=min(res, min(r, min(c, min(n-1-r, m-1-c))));
        if (res==0) break;
        if (v[r][c]==k) continue;

        for(int i=0; i<4; ++i) {
            int nr=r+dr[i];
            int nc=c+dc[i];
            if (nr<0 or nr>=n or nc<0 or nc>=m) continue;
            if (a[nr][nc]=='#') continue;
            if (~v[nr][nc]) continue;
            q.emplace(nr, nc);
            v[nr][nc]=v[r][c]+1;
        }
    }
    cout<<(res+k-1)/k+1<<endl;
}