// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define sz(x) int(x.size())
using db = double;
using ll = long long;
template<class T> using V = vector<T>;
using vd = V<db>;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vd A(N), B(N); for(int i = 0; i < N; i++) {
cin >> A[i] >> B[i];
}
sort(rbegin(A), rend(A)); sort(rbegin(B), rend(B));
for(int i = 1; i < N; i++) A[i] += A[i - 1];
for(int i = 1; i < N; i++) B[i] += B[i - 1];
// for(auto& x : A) cout << x << " ";
// cout << endl;
// for(auto& x : B) cout << x << " ";
// cout << endl;
db ans = 0;
for(int x = 2; x <= 2 * N; x++) {
// cout << x << " ===> " << endl;
// take x odds
auto F = [&](int i) {
if (i <= 0) return db(0);
if (x - i <= 0) return db(0);
return min(A[i - 1], B[x - 1 - i]) - x;
};
int lo = max(1, x - N), hi = min(x - 1, N);
while(lo < hi) {
int mid = (lo + hi + 1) / 2;
if (A[mid - 1] < B[x - 1 - mid]) lo = mid;
else hi = mid - 1;
}
ans = max(ans, F(lo - 1));
ans = max(ans, F(lo));
ans = max(ans, F(lo + 1));
// maximize min_{1 <= i <= x}(A[i - 1] - x, B[x - i] - x)
}
printf("%.4lf", ans);
exit(0-0);
}
