#include <bits/stdc++.h>
using namespace std;
using ld = long double;
const ld INF = 1e30L;
struct Fenwick {
int n;
vector<long long> cnt; // counts (integral)
vector<ld> sum; // sums (long double)
Fenwick(int n=0){ init(n); }
void init(int n_){
n = n_;
cnt.assign(n+1,0);
sum.assign(n+1,0);
}
void add(int i, long long dc, ld ds){
for(; i<=n; i += i&-i){
cnt[i] += dc;
sum[i] += ds;
}
}
long long pref_cnt(int i){
long long r=0;
for(; i>0; i-=i&-i) r+=cnt[i];
return r;
}
ld pref_sum(int i){
ld r=0;
for(; i>0; i-=i&-i) r+=sum[i];
return r;
}
// find smallest idx such that pref_cnt(idx) >= need (1-based), or n+1 if not enough
int find_by_count(long long need){
if(need <= 0) return 0;
long long cur = 0;
int pos = 0;
int LOG = 1;
while((1<<LOG) <= n) ++LOG;
for(int k = LOG; k >= 0; --k){
int nxt = pos + (1<<k);
if(nxt <= n && cur + cnt[nxt] < need){
pos = nxt;
cur += cnt[nxt];
}
}
return pos+1; // pos is last with sum < need
}
long long total_cnt(){ return pref_cnt(n); }
ld total_sum(){ return pref_sum(n); }
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, K;
if(!(cin >> n >> K)) return 0;
vector<ld> Aorig(n);
vector<ld> Borig(n);
for(int i=0;i<n;i++){
long long Ai; long long Bi;
cin >> Ai >> Bi;
Aorig[i] = (ld)Ai;
if(Bi == -1) Borig[i] = INF;
else Borig[i] = (ld)Bi;
}
// pair by B then A
vector<pair<ld,ld>> v;
v.reserve(n);
for(int i=0;i<n;i++) v.push_back({Borig[i], Aorig[i]});
sort(v.begin(), v.end()); // rosnaco po B
vector<ld> B(n+1), A(n+1);
for(int i=0;i<n;i++){
B[i+1] = v[i].first;
A[i+1] = v[i].second;
}
// prefix sum of B_j / j
vector<ld> prefBdiv(n+1, 0.0L);
for(int j=1;j<=n;j++){
prefBdiv[j] = prefBdiv[j-1] + B[j] / (ld)j;
}
// prepare compression for A values (we will only ever store A[1..n])
vector<ld> allA;
allA.reserve(n);
for(int i=1;i<=n;i++) allA.push_back(A[i]);
sort(allA.begin(), allA.end());
allA.erase(unique(allA.begin(), allA.end(), [](ld x, ld y){ return fabsl(x-y) < 1e-12L; }), allA.end());
int M = (int)allA.size();
auto get_rank = [&](ld x)->int{
// 1-based rank of x in allA
int pos = int(lower_bound(allA.begin(), allA.end(), x - 1e-18L) - allA.begin());
return pos+1;
};
// build Fenwick with all A initially (we will remove helpers as m grows)
Fenwick bit(M);
for(int i=1;i<=n;i++){
int r = get_rank(A[i]);
bit.add(r, 1, A[i]);
}
// count how many helpers are actually available (B < INF)
int num_helpers = 0;
for(int i=1;i<=n;i++) if(B[i] < INF/2) num_helpers++;
int max_m = min(num_helpers, max(0, K-1)); // nie ma sensu brać więcej niż K-1 pomagierów
ld answer = 1e100L;
// dla m = 0: nie usuwamy nic, czas zakupu = 0, potem wybieramy K smallest A z całego zbioru
for(int m = 0; m <= max_m; ++m){
// jeżeli m>0: usuwamy element A[m] (bo ten stan stał się pomagierem)
if(m > 0){
int rnk = get_rank(A[m]); // m-th after sort by B
bit.add(rnk, -1, -A[m]);
}
// czas zakupu pierwszych m pomagierów:
ld time_buy = prefBdiv[m]; // sum_{j=1..m} B[j] / j
int votes_from_helpers = m; // bo dla wybranych pomagierów Ai <= Bi
int need = max(0, K - votes_from_helpers);
if(need == 0){
answer = min(answer, time_buy);
continue;
}
if((long long)bit.total_cnt() < need) continue; // za malo pozostalych stanow -> niemozliwe
// znajdź sumę najmniejszych 'need' A w bieżącym zbiorze (pozostałe nie-kupione)
int idx = bit.find_by_count(need);
// sum do idx-1 w pełni + ewentualnie kilka z idx jeśli duplikaty
long long cnt_before = bit.pref_cnt(idx-1);
ld sum_before = bit.pref_sum(idx-1);
long long take_from_idx = need - cnt_before;
ld val_idx = allA[idx-1]; // allA jest 0-based, idx jest 1-based
ld sum_need = sum_before + val_idx * (ld)take_from_idx;
ld time_votes = sum_need / (ld)(m+1);
ld total_time = time_buy + time_votes;
answer = min(answer, total_time);
}
cout.setf(std::ios::fixed);
cout << setprecision(9) << (double)answer << "\n";
return 0;
}
