/*
sort by b from small to large
optimal construction can be split into two halves
- on the left, all the things are taken
- on the right, only a's are taken
fix the number of b's, let this be t
f(i,j) = min time to use prefix i with j b's
f(i,j) = min(f(i-1,j) + a[i] / t, f(i-1,j-1) + b[i] / t)
now for the suffixes, we need to pick k-i a's at the end,
we must pick the ones with the least a[j] value. this part
can be done in O(n log^2 n)
overall the time complexity is O(n^3)
*/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 0
#endif
const int N = 505;
const double INF = 1e15;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int n, k;
cin >> n >> k;
static array<int, 2> p[N];
for (int i = 0; i < n; i++) {
int a, b;
cin >> a >> b;
if (b == -1)
b = 1e9;
p[i] = {b, a};
}
sort(p, p + n);
static int g[N][N];
for (int i = 0; i < n; i++) {
vector<int> t;
for (int j = i; j < n; j++)
t.push_back(p[j][1]);
sort(t.begin(), t.end());
g[i][0] = 0;
for (int j = 1; j <= (int) t.size(); j++)
g[i][j] = g[i][j - 1] + t[j - 1];
}
double ans = INF;
for (int t = 0; t <= k; t++) {
static double f[N];
for (int i = 0; i <= k; i++)
f[i] = INF;
f[0] = 0;
for (int i = 0; i < n; i++) {
for (int j = k; j >= 0; j--)
f[j] = min(f[j] + (double) p[i][1] / (t + 1), j == 0 ? INF : f[j - 1] + (double) p[i][0] / j);
if (k - i - 1 >= 0)
ans = min(ans, f[t] + (double) g[i + 1][k - i - 1] / (t + 1));
}
}
cout << fixed << setprecision(12) << ans << '\n';
return 0;
}
