Problem: While Grisha was celebrating New Year with Ded Moroz, Misha gifted Sasha a small rectangular pond of size n × m, divided into cells of size 1 × 1, inhabited by tiny evil fishes (no more than one fish per cell, otherwise they'll strife!).

The gift bundle also includes a square scoop of size r × r, designed for fishing. If the lower-left corner of the scoop-net is located at cell (x, y), all fishes inside the square (x, y)...(x + r - 1, y + r - 1) get caught. Note that the scoop-net should lie completely inside the pond when used.

Unfortunately, Sasha is not that skilled in fishing and hence throws the scoop randomly. In order to not frustrate Sasha, Misha decided to release k fishes into the empty pond in such a way that the expected value of the number of caught fishes is as high as possible. Help Misha! In other words, put k fishes in the pond into distinct cells in such a way that when the scoop-net is placed into a random position among (n - r + 1)·(m - r + 1) possible positions, the average number of caught fishes is as high as possible.

Input

The only line contains four integers n, m, r, k (1 ≤ n, m ≤ 105, 1 ≤ r ≤ min(n, m), 1 ≤ k ≤ min(n·m, 105)).

Output

Print a single number — the maximum possible expected number of caught fishes.

You answer is considered correct, is its absolute or relative error does not exceed 10 - 9. Namely, let your answer be a, and the jury's answer be b. Your answer is considered correct, if <image>.

Examples

Input

3 3 2 3


Output

2.0000000000


Input

12 17 9 40


Output

32.8333333333

Note

In the first example you can put the fishes in cells (2, 1), (2, 2), (2, 3). In this case, for any of four possible positions of the scoop-net (highlighted with light green), the number of fishes inside is equal to two, and so is the expected value.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
void __print(long long first) { cerr << first; }
void __print(long first) { cerr << first; }
void __print(unsigned first) { cerr << first; }
void __print(unsigned long first) { cerr << first; }
void __print(unsigned long long first) { cerr << first; }
void __print(float first) { cerr << first; }
void __print(double first) { cerr << first; }
void __print(long double first) { cerr << first; }
void __print(char first) { cerr << '\'' << first << '\''; }
void __print(const char *first) { cerr << '\"' << first << '\"'; }
void __print(const string &first) { cerr << '\"' << first << '\"'; }
void __print(bool first) { cerr << (first ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &first) {
  cerr << '{';
  __print(first.first);
  cerr << ',';
  __print(first.second);
  cerr << '}';
}
template <typename T>
void __print(const T &first) {
  long long f = 0;
  cerr << '{';
  for (auto &i : first) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
std::mt19937 rng(
    (long long)std::chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1 << 20;
long long n, m, r, k;
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, 1, -1};
map<pair<long long, long long>, long long> vis;
long long get(pair<long long, long long> a) {
  long long first = a.first, second = a.second;
  return (min(first, n - r + 1) - max(first - r + 1, 1ll) + 1) *
         (min(second, m - r + 1) - max(second - r + 1, 1ll) + 1);
}
bool ok(long long first, long long second) {
  return (first > 0 and second > 0 and first <= n and second <= m);
}
struct cmp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) {
    long long fa = get(a);
    long long fb = get(b);
    if (fa == fb) return a > b;
    return fa > fb;
  }
};
set<pair<long long, long long>, cmp> s;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  cin >> n >> m >> r >> k;
  pair<long long, long long> d = {(n + 1) / 2, (m + 1) / 2};
  s.insert(d);
  vis[d] = 1;
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    auto p = *s.begin();
    s.erase(p);
    long long first = p.first, second = p.second;
    ans += get(p);
    for (long long dir = 0; dir < 4; dir++) {
      long long nx = first + dx[dir];
      long long ny = second + dy[dir];
      if (ok(nx, ny) and !vis[{nx, ny}]) {
        s.insert({nx, ny});
        vis[{nx, ny}] = 1;
      }
    }
  }
  long double res = 1.0 * ans / (1.0 * (n - r + 1) * (m - r + 1));
  cout << res;
  return 0;
}