#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
inline bool mina(A &first, B second) {
  return (first > second) ? (first = second, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &first, B second) {
  return (first < second) ? (first = second, 1) : 0;
}
template <class T>
class RangeMinimumQuery {
 public:
  vector<vector<int> > M;
  vector<T> A;
  int N;
  RangeMinimumQuery(T in[], int n) {
    N = n;
    M = vector<vector<int> >(N + 5);
    A = vector<T>(N + 5);
    int b;
    for (b = 0; 1 << b <= N; b++)
      ;
    for (int i = 0; i < N + 5; i++) M[i] = vector<int>(b);
    for (int i = 0; i < N; i++) {
      M[i][0] = i;
      A[i] = in[i];
    }
    for (int j = 1; 1 << j <= N; j++) {
      for (int i = 0; i + (1 << j) - 1 < N; i++)
        if (A[M[i][j - 1]] < A[M[i + (1 << (j - 1))][j - 1]])
          M[i][j] = M[i][j - 1];
        else
          M[i][j] = M[i + (1 << (j - 1))][j - 1];
    }
  }
  T get_min(int left, int right) {
    int num = right - left + 1;
    int e = 1, k = 0;
    while (e * 2 <= num) e *= 2, k++;
    int temp = right - e + 1;
    if (A[M[left][k]] <= A[M[temp][k]])
      return A[M[left][k]];
    else
      return A[M[temp][k]];
  }
};
int N, K;
int V[(1000005)];
int C[(1000005)];
int next_greater[(1000005)];
pair<int, int> potential[(1000005)];
int final_potential[(1000005)];
int main() {
  scanf("%d %d", &N, &K);
  for (int(i) = 0; (i) < (N); ++(i)) scanf("%d", &V[i]);
  for (int(i) = 0; (i) < (N); ++(i)) scanf("%d", &C[i]);
  RangeMinimumQuery<int> rc(C, N);
  next_greater[N - 1] = N;
  for (int i = N - 2; i >= 0; i--) {
    int now = i + 1;
    while (now < N && V[now] <= V[i]) {
      now = next_greater[now];
    }
    next_greater[i] = now;
  }
  potential[N - 1] = pair<int, int>(V[N - 1], C[N - 1]);
  for (int i = N - 2; i >= 0; i--) {
    if (next_greater[i] == N) {
      potential[i] = pair<int, int>(V[i], C[i]);
    } else {
      int ng = next_greater[i];
      int pp = min(rc.get_min(i, ng), potential[ng].second);
      if (min(100 * potential[ng].first, pp) > min(100 * V[i], C[i])) {
        potential[i] = pair<int, int>(potential[ng].first, pp);
      } else {
        potential[i] = pair<int, int>(V[i], C[i]);
      }
    }
  }
  for (int(i) = 0; (i) < (N); ++(i)) {
    final_potential[i] = min(100 * potential[i].first, potential[i].second);
  }
  sort(final_potential, final_potential + N);
  double ans = 0.0;
  double p1 = 1.0 * K / N;
  int a = N - K;
  int b = N - 1;
  for (int i = 0; N - i - 1 >= K - 1; i++) {
    ans += final_potential[i] * p1;
    p1 *= a;
    p1 /= b;
    a--;
    b--;
  }
  printf("%.9lf\n", ans);
  return 0;
}
