Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3 3 3
#.#
#S.
###

Sample Output 1

1

Takahashi can move to room (1,2) in one cast.

Sample Input 2

3 3 3
###
#S#
###

Sample Output 2

2

Takahashi cannot move in the first cast, but can unlock room (1,2).
Then, he can move to room (1,2) in the next cast, achieving the objective in two casts.

Sample Input 3

7 7 2
#######
#######
##...##
###S###
##.#.##
###.###
#######

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

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



typedef unsigned long ul;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<ll> vint;
typedef vector< vector<ll> > vvint;
typedef vector< vector< vector<ll> > > vvvint;
typedef vector<string> vstring;
typedef vector< vector<string> > vvstring;
typedef vector<char> vchar;
typedef vector< vector<char> > vvchar;
typedef vector<long double> vdouble;
typedef vector< vector<long double> > vvdouble;
typedef vector< vector< vector<long double> > > vvvdouble;
typedef pair<ll,ll> pint;
typedef vector<pint> vpint;
typedef vector<bool> vbool;

#define rep(i,n) for(ll i=0;i<n;i++)
#define repf(i,f,n) for(ll i=f;i<n;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ALL(obj) (obj).begin(), (obj).end()
// #define LLONG_MAX 9223372036854775806
#define vmax(vec) *max_element(vec.begin(), vec.end())
#define vmin(vec) *min_element(vec.begin(), vec.end())
#define vsort(vec) sort(vec.begin(), vec.end())
#define vsortgr(vec) sort(vec.begin(), vec.end(), greater<ll>())
#define MOD 1000000007
// #define MOD 998244353
// #define MOD LLONG_MAX
const double PI=3.14159265358979323846;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }




int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0,};

template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;

ll gcd(ll a, ll b)
{
    if (a < b) {
        a ^= b;
        b ^= a;
        a ^= b;
    }

    return b ? gcd(b, a % b) : a;
}

ll lcm(int a, ll b)
{
    return a * b / gcd(a, b);
}




// 繰り返し二乗法
ll power(ll a, ll b){
  if(a==1)return a;
  // if(a==0)re
  ll res=1;
  while(b>0){
    if(b&1) res=res*a%MOD;
    a=a*a%MOD;
    b>>=1;
  }
  return res;
}

const int MAX=1000000;

ll fact[MAX], fact_inv[MAX];

void init_fact(ll n){
  fact[0]=1;
  // 階乗の計算
  rep(i, n) fact[i+1]=fact[i]*(i+1)%MOD;
  fact_inv[n]=power(fact[n], MOD-2);
  // 逆元の計算
  for(ll i=n-1;i>=0;i--) fact_inv[i]=fact_inv[i+1]*(i+1)%MOD;
}

ll comb(ll n, ll r){
  return (fact[n]*fact_inv[r])%MOD*fact_inv[n-r]%MOD;
}

ll perm(ll n, ll r){
  return (fact[n]*fact_inv[n-r])%MOD;
}


struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<ll> num;
    // vector<ll> dist; // rootまでの距離

    UnionFind(ll N) : par(N), num(N){ //最初は全てが根であるとして初期化
      for(ll i = 0; i < N; i++) par[i] = i;
      for(ll i = 0; i < N; i++) num[i] = 1;
      // for(ll i = 0; i < N; i++) dist[i] = 0;

    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
        num[ry] = num[rx] + num[ry];
    }



    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }

    ll size(ll x){
      return num[root(x)];
    }


};

vint divisor(ll n){ // nの約数
  vint ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i == 0){
      ret.pb(i);
      if(i*i!=n) ret.pb(n/i);
    }
  }
  vsort(ret);
  return ret;
}



ll get_digit_in(ll n, ll i){ // i桁目の数字を得る。
  for(ll j=0;j<i-1;j++){
    n/=10;
  }
  return n%10;
}

ll get_digit(ll n){
  ll rtn=0;
  while(n>0){
    n/=10;
    rtn++;
  }
  return rtn;
}

map<ll,ll> prime_factor(ll n){
  map<ll,ll> rtn;
  for(ll i=2;i*i<=n;i++){
    while(n%i == 0){
      rtn[i]++;
      n /= i;
    }
  }
  if(n!=1) rtn[n] = 1;
  return rtn;
}

// nPnの順列に対して処理を実行する
void foreach_permutation(ll n, function<void(ll *)> f) {
  ll indexes[n];
  for (ll i = 0; i < n; i++) indexes[i] = i;
  do {
    f(indexes);
  } while (std::next_permutation(indexes, indexes + n));
}

void recursive_comb(ll *indexes, ll s, ll rest, function<void(ll *)> f) {
  if (rest == 0) {
    f(indexes);
  } else {
    if (s < 0) return;
    recursive_comb(indexes, s - 1, rest, f);
    indexes[rest - 1] = s;
    recursive_comb(indexes, s - 1, rest - 1, f);
  }
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(ll n, ll k, function<void(ll *)> f) {
  ll indexes[k];
  recursive_comb(indexes, n - 1, k, f);
}

// nPkの順列に対して処理を実行する
void foreach_permutation(ll n, ll k, function<void(ll *)> f) {
  foreach_comb(n, k, [&](ll *c_indexes) {
    foreach_permutation(k, [&](ll *p_indexes) {
      ll indexes[k];
      for (ll i = 0; i < k; i++) {
        indexes[i] = c_indexes[p_indexes[i]];
      }
      f(indexes);
    });
  });
}

// ll arr[1000500];
// vint primes;

// //エラトステネスの篩
// void Eratosthenes(){
//   ll N = 1000500;
//  for(ll i = 0; i < N; i++){
//    arr[i] = 1;
//  }
//  for(ll i = 2; i < sqrt(N); i++){
//    if(arr[i]){
//      for(ll j = 0; i * (j + 2) < N; j++){
//        arr[i *(j + 2)] = 0;
//      }
//    }
//  }

//  for(ll i = 2; i < N; i++){
//    if(arr[i]){
//       primes.pb(i);
//      //cout << i << endl;
//    }
//  }
// }


void printv(vint &V){
  for(auto e:V)cout<<e<<" ";
  cout<<endl;
}

template< typename G >
struct DoublingLowestCommonAncestor {
  const int LOG;
  vector< int > dep;
  const G &g;
  vector< vector< int > > table;

  DoublingLowestCommonAncestor(const G &g) : g(g), dep(g.size()), LOG(32 - __builtin_clz(g.size())) {
    table.assign(LOG, vector< int >(g.size(), -1));
  }

  void dfs(int idx, int par, int d) {
    table[0][idx] = par;
    dep[idx] = d;
    for(auto &to : g[idx]) {
      if(to != par) dfs(to, idx, d + 1);
    }
  }

  void build() {
    dfs(0, -1, 0);
    for(int k = 0; k + 1 < LOG; k++) {
      for(int i = 0; i < table[k].size(); i++) {
        if(table[k][i] == -1) table[k + 1][i] = -1;
        else table[k + 1][i] = table[k][table[k][i]];
      }
    }
  }

  int query(int u, int v) {
    if(dep[u] > dep[v]) swap(u, v);
    for(int i = LOG - 1; i >= 0; i--) {
      if(((dep[v] - dep[u]) >> i) & 1) v = table[i][v];
    }
    if(u == v) return u;
    for(int i = LOG - 1; i >= 0; i--) {
      if(table[i][u] != table[i][v]) {
        u = table[i][u];
        v = table[i][v];
      }
    }
    return table[0][u];
  }
};


char A[1000][1000];
ll dist[1000][1000];
ll H,W,K; 

void bfs(ll i,ll j, ll d){
  if(i < 0 || i >= H || j < 0 || j >= W ) return;
  if(dist[i][j] <= K) return;
  if(A[i][j] == '#') return;
  dist[i][j] = d;

  bfs(i-1,j, d+1);
  bfs(i+1,j,d+1);
  bfs(i,j-1,d+1);
  bfs(i,j+1,d+1);
  return;
}


int main() {


  cout<<fixed<<setprecision(10);

  cin>>H>>W>>K;

  rep(i,H)rep(j,W) cin>>A[i][j];
  rep(i,H)rep(j,W) dist[i][j] = K+1;

  ll si, sj;
  si = -1;
  sj = -1;
  rep(i,H)rep(j,W){
    if(A[i][j] == 'S'){
      si = i; sj = j;
      break;
    }
  }

  assert(!(si == -1 && sj == -1));

  bfs(si, sj, 0);

  // rep(i,H){
  //   rep(j,W){
  //     cout<<dist[i][j];
  //   }
  //   cout<<endl;
  // }

  ll dmin = H*W;

  repf(i,1,H-1){
    // down
    ll cnt = 0;
    ll j = 0;
    while( (j <= W-1) && (dist[i][j] > K) ){
      cnt++;
      j++;
    }
    chmin(dmin, cnt);

    // up
    cnt = 0;
    j = W-1;
    while( (j >= 0) && (dist[i][j] > K) ){
      cnt++;
      j--;
    }

    chmin(dmin, cnt);
  }

  repf(j,1,W-1){
    // right
    ll cnt = 0;
    ll i = 0;
    while( (i <= H-1) && (dist[i][j] > K) ){
      cnt++;
      i++;
    }
    chmin(dmin, cnt);

    // left
    cnt = 0;
    i = H-1;
    while( (i >= 0) && (dist[i][j] > K) ){
      cnt++;
      i--;
    }

    chmin(dmin, cnt);
  }

  ll ans = 1 + (dmin + K - 1)/K;
  cout<<ans<<endl;
}




//