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 <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

char a[802][808];

int main() {
    int h, w, k;
    cin >> h >> w >> k;

    int i0, j0;
    for (int i = 1; i <= h; i++) {
        cin >> a[i];
        for (int j = 0; j < w; j++) {
            if (a[i][j] == 'S') i0 = i, j0 = j, a[i][j] = '.';
        }
    }

    queue<pair<int, int>> q;
    q.push({ i0, j0 });
    int d = 800;

    for (int l = 0; l <= k; l++) {
        int n = (int)q.size();
        for (int m = 0; m < n; m++) {
            auto p = q.front(); q.pop();
            int i = p.first, j = p.second;
            if (a[i][j] != '.') continue;
            a[i][j] = '+';
            d = min(d, i - 1);
            d = min(d, h - i);
            d = min(d, j);
            d = min(d, w - 1 - j);
            q.push({ i - 1, j });
            q.push({ i, j - 1 });
            q.push({ i, j + 1 });
            q.push({ i + 1, j });
        }
    }

    cout << (d + k - 1) / k + 1 << endl;

    return 0;
}