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;

int main(int argc, const char *argv[]) {
  int n, m, k;
  cin >> n >> m >> k;
  vector<string> a(n);
  int sx, sy;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'S') {
        sx = i, sy = j;
      }
    }
  }
  int dx[] = {-1, 0, 0, 1};
  int dy[] = {0, -1, 1, 0};
  queue<pair<int, int>> q;
  vector<vector<int>> dis(n, vector<int>(m, -1));
  dis[sx][sy] = 0;
  q.push({sx, sy});
  while (!q.empty()) {
    auto u = q.front();
    q.pop();
    int x = u.first, y = u.second;
    for (int i = 0; i < 4; i++) {
      int tx = x + dx[i], ty = y + dy[i];
      if (tx >= 0 && tx < n && ty >= 0 && ty < m && a[tx][ty] == '.' && dis[tx][ty] == -1) {
        dis[tx][ty] = dis[x][y] + 1;
        q.push({tx, ty});
      }
    }
  }
  int ans = 1e9;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (dis[i][j] != -1 && dis[i][j] <= k) {
        int minn = min(i, n - 1 - i);
        minn = min(minn, min(j, m - 1 - j));
        ans = min(ans, minn / k + (minn % k > 0));
      }
    }
  }
  cout << ans + 1 << endl;
  return 0;
}