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
#include<bits/stdc++.h>
using namespace std;

int main() {
  int H, W, K;
  cin >> H >> W >> K;
  vector<string> A(H);
  for(auto& i: A) cin >> i;

  const int dy[] = {0, 0, 1, -1};
  const int dx[] = {1, -1, 0, 0};
  int sy, sx;
  for(auto i=0; i<H; ++i) for(auto j=0; j<W; ++j) if(A[i][j]=='S') sy= i, sx = j;

  auto res = H*W;

  using P = tuple<int, int, int>;
  queue<P> q;
  q.emplace(0, sy, sx);
  while(!q.empty()) {
    int d, y, x;
    tie(d, y, x) = q.front();
    q.pop();
    auto r = min({y, H-1-y, x, W-1-x});
    res = min(res, 1 + (r+K-1)/K);
    if(d == K) continue;
    for(auto i=0; i<4; ++i) {
      auto ny = y + dy[i];
      auto nx = x + dx[i];
      if(ny<0 || H<=ny) continue;
      if(nx<0 || W<=nx) continue;
      if(A[ny][nx]!='.') continue;
      A[ny][nx] = '#';
      q.emplace(d+1, ny, nx);
    }
  }

  cout << res << endl;
}
