#include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct Fenwick {
int n;
vector<long double> bitSum;
vector<int> bitCnt;
Fenwick(int n=0){ init(n); }
void init(int n_){
n = n_;
bitSum.assign(n+1, 0.0L);
bitCnt.assign(n+1, 0);
}
void add(int idx, long double val, int cnt){
for(int i = idx; i <= n; i += i & -i){
bitSum[i] += val;
bitCnt[i] += cnt;
}
}
// prefix sum, prefix count
long double sumPrefix(int idx){
long double s = 0;
for(int i = idx; i > 0; i -= i & -i) s += bitSum[i];
return s;
}
int cntPrefix(int idx){
int s = 0;
for(int i = idx; i > 0; i -= i & -i) s += bitCnt[i];
return s;
}
// find smallest position p such that cntPrefix(p) >= k. assume k >= 1 and k <= total count
int findByCount(int k){
int pos = 0;
int LOG = 1;
while((1<<LOG) <= n) ++LOG;
for(int pw = 1<<LOG; pw; pw >>= 1){
if(pos + pw <= n && bitCnt[pos + pw] < k){
k -= bitCnt[pos + pw];
pos += pw;
}
}
return pos + 1;
}
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N, K;
if(!(cin >> N >> K)) return 0;
vector<long double> A(N);
vector<long double> B(N);
for(int i=0;i<N;++i){
ll ai, bi;
cin >> ai >> bi;
A[i] = (long double)ai;
B[i] = (long double)bi; // bi == -1 means no collaborator
}
// collect collaborator candidates
vector<pair<long double,int>> coll; // (Bi, idx)
for(int i=0;i<N;++i) if(B[i] >= 0) coll.emplace_back(B[i], i);
sort(coll.begin(), coll.end(), [](auto &p1, auto &p2){
if(p1.first != p2.first) return p1.first < p2.first;
return p1.second < p2.second;
});
// prepare Ai sorted list (value, idx)
vector<pair<long double,int>> allA;
allA.reserve(N);
for(int i=0;i<N;++i) allA.emplace_back(A[i], i);
sort(allA.begin(), allA.end());
// Fenwick over sorted by Ai positions: initially include all states
Fenwick fw(N);
// position map: idx -> position in sorted allA (1-based)
vector<int> pos(N);
for(int i=0;i<N;++i){
pos[allA[i].second] = i+1;
}
for(int i=0;i<N;++i){
fw.add(i+1, (long double)allA[i].first, 1);
}
long double ans = 1e100L;
int maxColl = (int)coll.size();
int limit = min(K, maxColl);
// We will iterate i = number of collaborators we will obtain (0..limit)
// As we increase i, we remove that state's Ai from BIT (since it's already "used" and counted among votes)
long double t0 = 0.0L; // prefix time to get i collaborators
for(int i = 0; i <= limit; ++i){
int haveVotesFromColl = i; // each collaborator state gives vote (Bi >= Ai guaranteed)
int need = K - haveVotesFromColl;
if(need <= 0){
// already have >= K votes
ans = min(ans, t0);
} else {
// check how many available states remain in BIT
int available = fw.cntPrefix(N);
if(available >= need){
// get sum of smallest 'need' Ai among available
int posK = fw.findByCount(need);
int cntBefore = fw.cntPrefix(posK-1);
long double sumBefore = fw.sumPrefix(posK-1);
int takeAtPos = need - cntBefore;
long double valAtPos = allA[posK-1].first;
long double sumNeed = sumBefore + valAtPos * (long double)takeAtPos;
long double totalTime = t0 + sumNeed / (1.0L + (long double)i);
if(totalTime < ans) ans = totalTime;
}
}
// prepare for next i: if i == limit we stop; otherwise include next collaborator (remove its Ai from BIT and add time to t0)
if(i == limit) break;
// collaborator to be obtained is coll[i] (0-based)
long double Bi = coll[i].first;
int idx = coll[i].second;
// time to get this collaborator when currently have (1 + i) speakers: we must spend Bi / (1 + i) additional time
// BUT we must consider cumulative effect: previously t0 already sum for earlier collaborators, so add Bi / (1+i)
t0 += Bi / (1.0L + (long double)i);
// This collaborator's Ai is already satisfied (Bi >= Ai), so remove its Ai from the pool if still present.
int p = pos[idx];
// check if still present (count at p might be 1)
// safe to always attempt to remove; but need to ensure we only remove once
// We'll query count prefix and single index count:
// get count at single index by cntPrefix(p) - cntPrefix(p-1)
int cntAt = fw.cntPrefix(p) - fw.cntPrefix(p-1);
if(cntAt > 0){
fw.add(p, - (long double)allA[p-1].first, -1);
}
}
// print with required format; error tolerance <= 0.01
cout.setf(std::ios::fixed); cout<<setprecision(12);
double out = (double)ans;
cout << out << "\n";
return 0;
}
