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 H, W, K;
#define MAX_H 800
#define MAX_W 800
int SI, SJ;
bool U[MAX_H][MAX_W];
int DI[4] = {-1, 0, 1, 0};
int DJ[4] = {0, -1, 0, 1};
int R = INT_MAX;

int main() {
  cin >> H >> W >> K;
  for(int i = 0; i < H; ++i) for(int j = 0; j < W; ++j) {
    char c; cin >> c;
    U[i][j] = c != '.';
    if(c == 'S') { SI = i; SJ = j; }
  }
  queue<tuple<int, int, int>> q; q.push(make_tuple(SI, SJ, 0));
  while(!q.empty()) {
    auto t = q.front(); q.pop(); int i = get<0>(t), j = get<1>(t), d = get<2>(t);
    R = min({R, i, H - 1 - i, j, W - 1 - j});
    if(d == K) continue;
    for(int x = 0; x < 4; ++x) {
      int di = DI[x], dj = DJ[x];
      if(i + di < 0 || i + di >= H || j + dj < 0 || j + dj >= W) continue;
      if(U[i + di][j + dj]) continue;
      U[i + di][j + dj] = true; q.push(make_tuple(i + di, j + dj, d + 1));
    }
  }
  cout << 1 + (R + K - 1) / K << endl;
  return 0;
}