// Solution to: Easter Eggs
// By: Raymond van Bommel
// Wrong answer by greedily removing vertices with most neighbours

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <climits>
#include <cassert>

#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <string>

#include <iostream>
#include <sstream>

#include <utility>
#include <functional>
#include <limits>
#include <numeric>
#include <algorithm> 

using namespace std;

#define MaxN 250
int N, B, R;
double Bx[MaxN], By[MaxN], Rx[MaxN], Ry[MaxN];
bool adj[2*MaxN][2*MaxN];
int degree[2*MaxN];

double distance(int i, int j) {
	return sqrt( (Bx[i] - Rx[j])*(Bx[i] - Rx[j]) + (By[i] - Ry[j])*(By[i] - Ry[j]) );
}



bool poss(double D) {
	// Check if N eggs can be put, such that the distance between blue and red is at least D.
	for (int i = 0; i < B+R; i++) {	// Begin initialisation
		degree[i] = 0;
		for (int j = 0; j < B+R; j++)
			adj[i][j] = false;
	}
	
	for (int i = 0; i < B; i++)
		for (int j = 0; j < R; j++)
			if (distance(i,j) < D) {
				adj[i][j+B] = true;
				adj[j+B][i] = true;
				degree[i]++;
				degree[j+B]++;
			}
			
	priority_queue < pair<int, int> > q;
	for (int i = 0; i < B+R; i++)
		q.push(make_pair(degree[i], i));	// End initialisation
	
	int removed = 0;
	while (removed < B+R-N) {	// Repeatedly look for the vertex with most neighbours
		pair<int,int> p = q.top();
		q.pop();
		if (degree[p.second] != p.first) { // If degree has changed, do not consider and put in the queue again.
			q.push(make_pair(degree[p.second], p.second));
		} else {
			removed++;	// Remove the vertex, all adjacency relations, and decrease the degrees of the neighbours.
			int i = p.second;
			for (int j = 0; j < B+R; j++) {
				if (adj[i][j]) {
					degree[j]--;
					adj[i][j] = adj[j][i] = false;
				}
			}	
		}
	}
	
	for (int i = 0; i < B+R; i++)
	for (int j = 0; j < B+R; j++) { // Check if the graph has any edges left, if so, return false.
		if ( (i != j) && (adj[i][j]) ) {
			return false;
		}
	} // Otherwise return 0.
	return true;	
}


int main () {
	cin >> N >> B >> R;
	assert(N <= 250);
	assert(B < N);
	assert(R < N);
	for (int i = 0; i < B; i++) {
		cin >> Bx[i] >> By[i];
		assert(abs(Bx[i]) <= 10000);
		assert(abs(By[i]) <= 10000);	
	}
	for (int j = 0; j < R; j++) {
		cin >> Rx[j] >> Ry[j];
		assert(abs(Rx[j]) <= 10000);	
		assert(abs(Ry[j]) <= 10000);
	}
	
	// Binary search the minimal distance between blue and red egg.
	double hi = sqrt(2.0)*2000000.1;
	double lo = 0.0;
	while (hi - lo > 1E-7) {
		double mid = (hi+lo)/2.0;
		(poss(mid) ? lo : hi) = mid;
	}
	cout.precision(6);
	cout << fixed << hi << '\n';
	return 0;
}