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 <string>
#include <queue>
using namespace std;
typedef pair<int, int> P;

const int INF = 10000000;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

int d[803][803];

int main()
{
    int h, w, k;
    cin >> h >> w >> k;
    string a[803];
    for(int i = 0; i < h; i++){
        cin >> a[i];
    }
    queue<P> que;
    for(int i = 0; i < h; i++){
        for(int j = 0; j < w; j++){
            if(a[i][j] == 'S') que.push(P(i, j));
            else d[i][j] = -1;
        }
    }
    int ans = INF;
    while(que.size()){
        P p = que.front();
        que.pop();
        int x = p.first, y = p.second;
        for(int i = 0; i < 4; i++){
            int u = x + dx[i], v = y + dy[i];
            if(u >= 0 && u < h && v >= 0 && v < w && a[u][v] != '#' && d[u][v] == -1){
                d[u][v] = d[x][y] + 1;
                que.push(P(u, v));
            }
        }
        if(d[x][y] <= k){
            ans = min(ans, min(min(x + k - 1, y + k - 1), min(h - x + k - 2, w - y + k - 2)) / k + 1);
        }
    }
    cout << ans << endl;
}