#include <bits/stdc++.h>
namespace chtholly {
char buf[1 << 23], *buf1 = buf, *buf2 = buf;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= c == '-';
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ '0');
  return f ? x : -x;
}
template <typename mitsuha>
inline bool read(mitsuha &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c) && ~c; c = getchar()) f ^= c == '-';
  if (!~c) return 0;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ '0');
  return x = f ? x : -x, 1;
}
template <typename mitsuha>
inline int write(mitsuha x) {
  if (!x) return 0 & putchar(48);
  if (x < 0) putchar('-'), x = -x;
  int bit[20], i, p = 0;
  for (; x; x /= 10) bit[++p] = x % 10;
  for (i = p; i; --i) putchar(bit[i] + 48);
  return 0;
}
inline char fuhao() {
  char c = getchar();
  for (; isspace(c); c = getchar())
    ;
  return c;
}
}  // namespace chtholly
using namespace chtholly;
using namespace std;
const int z = 233;
const double e = 2.7182818284590452353602874713527;
int n, x[z], y[z], cx, cy, k;
double fail, dp[z][z];
double dis(int a, int b, int k) {
  return sqrt((a - x[k]) * (a - x[k]) + (b - y[k]) * (b - y[k]));
}
bool akioi(double br) {
  double t;
  int i, j;
  memset(dp, 0, sizeof dp);
  for (**dp = i = 1; i <= n; ++i)
    for (dp[i][0] = dp[i - 1][0] * ((t = dis(cx, cy, i)) <= br
                                        ? 0
                                        : (1 - pow(e, 1 - t * t / br / br))),
        j = 1;
         j <= n; ++j)
      dp[i][j] = (t = dis(cx, cy, i)) <= br
                     ? dp[i - 1][j - 1]
                     : dp[i - 1][j] * (1 - pow(e, 1 - t * t / br / br)) +
                           dp[i - 1][j - 1] * pow(e, 1 - t * t / br / br);
  double llx = 0;
  for (i = 0; i < k; ++i) llx += dp[n][i];
  return llx > fail;
}
int main() {
  read(n);
  int i, t;
  k = read(), fail = 1.0 * read() / 1000;
  cx = read(), cy = read();
  for (i = 1; i <= n; ++i) x[i] = read(), y[i] = read();
  double l = 0, r = 1e5, mid;
  for (t = 100; t--; akioi(mid) ? l = mid : r = mid) mid = (l + r) / 2;
  printf("%.12lf\n", r);
}
