#include <bits/stdc++.h>
using namespace std;
const int maxn = 50005;
const long double eps = 1e-10;
inline int gi() {
  char c = getchar();
  bool f = 1;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') c = getchar(), f = 0;
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return f ? sum : -sum;
}
int n, m, x, y, t, v[maxn], c[maxn * 2];
long double p, q, a[maxn], b[maxn];
pair<long double, int> s[maxn * 2];
struct P {
  long double x, y;
  inline P() {}
  inline P(long double x, long double y) : x(x), y(y) {}
  inline P &operator+=(P o) { return x += o.x, y += o.y, *this; }
  inline P &operator-=(P o) { return x -= o.x, y -= o.y, *this; }
  inline P &operator*=(long double o) { return x *= o, y *= o, *this; }
  inline P &operator/=(long double o) { return x /= o, y /= o, *this; }
  inline friend P operator+(P a, P b) { return a += b; }
  inline friend P operator-(P a, P b) { return a -= b; }
  inline friend P operator*(P a, long double b) { return a *= b; }
  inline friend P operator/(P a, long double b) { return a /= b; }
  inline friend bool operator<(P a, P b) {
    return fabs(a.x - b.x) < eps ? a.y < b.y : a.x < b.x;
  }
  inline friend bool operator==(P a, P b) { return !(a < b) && !(b < a); }
  inline friend long double operator*(P a, P b) {
    return a.x * b.x + a.y * b.y;
  }
  inline friend long double operator%(P a, P b) {
    return a.x * b.y - a.y * b.x;
  }
  inline friend long double operator^(P a, P b) {
    return a * b / a.l() / b.l();
  }
  inline long double a() { return atan2(y, x); }
  inline long double l() { return sqrt(*this * *this); }
  inline void r(long double o) {
    long double s = sin(o), c = cos(o), X = x * c - y * s, Y = x * s + y * c;
    x = X, y = Y;
  }
};
inline void add(int x) {
  while (x <= t) ++c[x], x += x & -x;
}
inline int ask(int x) {
  int k = 0;
  while (x) k += c[x], x -= x & -x;
  return k;
}
inline void solve(long double r) {
  t = 0;
  for (int i = 1; i <= n; i++) {
    long double a = ::a[i], b = ::b[i];
    long double d = 4 * a * a * b * b - 4 * (a * a + 1) * (b * b - r * r);
    if (d < eps) continue;
    d = sqrt(d);
    long double x1 = (-2 * a * b - d) / (2 * (a * a + 1));
    long double x2 = (-2 * a * b + d) / (2 * (a * a + 1));
    long double y1 = a * x1 + b, y2 = a * x2 + b;
    s[++t] = make_pair(P(x1, y1).a(), i), s[++t] = make_pair(P(x2, y2).a(), i);
    v[i] = 0;
  }
  sort(s + 1, s + t + 1);
  for (int i = t; i; i--)
    if (!v[s[i].second]) v[s[i].second] = i;
}
inline bool pd(long double r) {
  solve(r);
  for (int i = 1; i <= t; i++) c[i] = 0;
  long long ret = 0;
  for (int i = 1; i <= t; i++)
    if (v[s[i].second] != i) {
      ret += ask(v[s[i].second]) - ask(i - 1);
      add(v[s[i].second]);
    }
  return ret >= m;
}
inline long double calc(int i, int j) {
  long double x = (b[j] - b[i]) / (a[i] - a[j]), y = a[i] * x + b[i];
  return sqrt(x * x + y * y);
}
inline long double calc(long double r) {
  solve(r);
  set<pair<int, int> > st;
  long double ret = 0;
  int cnt = 0;
  for (int i = 1; i <= t; i++)
    if (v[s[i].second] != i) {
      auto it = st.upper_bound(make_pair(i + 1, 0));
      while (it != st.end() && (it->first) < v[s[i].second])
        ret += calc(s[i].second, it->second), ++it, ++cnt;
      st.insert(make_pair(v[s[i].second], s[i].second));
    }
  return ret + (m - cnt) * r;
}
int main() {
  n = gi();
  p = gi() / 1e3;
  q = gi() / 1e3;
  m = gi();
  for (int i = 1; i <= n; i++)
    a[i] = gi() / 1e3, b[i] = gi() / 1e3 + a[i] * p - q;
  long double l = 0, r = 1e10;
  for (int i = 1; i <= 120; i++) {
    long double mid = (l + r) / 2;
    if (pd(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%.9Lf\n", max(0.L, calc(r)));
  return 0;
}
