#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;
using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;
bool paircomp(const pair<ll, ll> &a, const pair<ll, ll> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll H, W, K;
  cin >> H >> W >> K;
  vvl A = vector<vector<ll>>(H, vector<ll>(W, 0));
  pair<ll, ll> start;
  for (ll i = 0; i < (H); i++)
    for (ll j = 0; j < (W); j++) {
      char s;
      cin >> s;
      if (s == '#')
        A[i][j] = 1;
      else if (s == 'S')
        start = make_pair(i, j);
    }
  ll ans = llINF;
  queue<pair<ll, pair<ll, ll>>> Q;
  Q.push(make_pair(0, start));
  while (!Q.empty()) {
    auto po = Q.front();
    Q.pop();
    ll count = po.first;
    if (po.second.first < 0 || po.second.first >= H || po.second.second < 0 ||
        po.second.second >= W || A[po.second.first][po.second.second] == 1)
      continue;
    ans = min(ans, min(min(po.second.first, po.second.second),
                       min(H - po.second.first, W - po.second.second)));
    if (count < K) {
      Q.push(make_pair(count + 1,
                       make_pair(po.second.first + 1, po.second.second)));
      Q.push(make_pair(count + 1,
                       make_pair(po.second.first - 1, po.second.second)));
      Q.push(make_pair(count + 1,
                       make_pair(po.second.first, po.second.second + 1)));
      Q.push(make_pair(count + 1,
                       make_pair(po.second.first, po.second.second - 1)));
    }
  }
  if (ans == 0)
    cout << 1 << endl;
  else {
    cout << 1 + ceil((long double)ans / K) << endl;
  }
  return 0;
}
