Problem: Gosha is hunting. His goal is to catch as many Pokemons as possible. Gosha has a Poke Balls and b Ultra Balls. There are n Pokemons. They are numbered 1 through n. Gosha knows that if he throws a Poke Ball at the i-th Pokemon he catches it with probability pi. If he throws an Ultra Ball at the i-th Pokemon he catches it with probability ui. He can throw at most one Ball of each type at any Pokemon.

The hunting proceeds as follows: at first, Gosha chooses no more than a Pokemons at which he will throw Poke Balls and no more than b Pokemons at which he will throw Ultra Balls. After that, he throws the chosen Balls at the chosen Pokemons. If he throws both Ultra Ball and Poke Ball at some Pokemon, he is caught if and only if he is caught by any of these Balls. The outcome of a throw doesn't depend on the other throws.

Gosha would like to know what is the expected number of the Pokemons he catches if he acts in an optimal way. In other words, he would like to know the maximum possible expected number of Pokemons can catch.

Input

The first line contains three integers n, a and b (2 ≤ n ≤ 2000, 0 ≤ a, b ≤ n) — the number of Pokemons, the number of Poke Balls and the number of Ultra Balls.

The second line contains n real values p1, p2, ..., pn (0 ≤ pi ≤ 1), where pi is the probability of catching the i-th Pokemon if Gosha throws a Poke Ball to it.

The third line contains n real values u1, u2, ..., un (0 ≤ ui ≤ 1), where ui is the probability of catching the i-th Pokemon if Gosha throws an Ultra Ball to it.

All the probabilities are given with exactly three digits after the decimal separator.

Output

Print the maximum possible expected number of Pokemons Gosha can catch. The answer is considered correct if it's absolute or relative error doesn't exceed 10 - 4.

Examples

Input

3 2 2
1.000 0.000 0.500
0.000 1.000 0.500


Output

2.75


Input

4 1 3
0.100 0.500 0.500 0.600
0.100 0.500 0.900 0.400


Output

2.16


Input

3 2 0
0.412 0.198 0.599
0.612 0.987 0.443


Output

1.011

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.AbstractQueue;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.AbstractCollection;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public static int n;
        public static int a;
        public static int b;
        public static double[] p;
        public static double[] u;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            n = in.nextInt();
            a = in.nextInt();
            b = in.nextInt();
            double sumall = 0;
            p = new double[n];
            for (int i = 0; i < n; i++) sumall += p[i] = in.nextDouble();
            u = new double[n];
            for (int j = 0; j < n; j++) sumall += u[j] = in.nextDouble();

            double ans = solve();
            out.printf("%.10f\n", -ans);
        }

        public static double solve() {
            int N = 2 + n + 2;
            List<TaskE.Edge>[] graph = new List[N];
            for (int i = 0; i < N; i++) graph[i] = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                addEdge(graph, 0, 2 + i, 1, -p[i]);
                addEdge(graph, 1, 2 + i, 1, -u[i]);
                addEdge(graph, 2 + i, N - 2, 1, 0);
                addEdge(graph, 2 + i, N - 2, 1, p[i] * u[i]);
            }
            addEdge(graph, N - 1, 0, a, 0);
            addEdge(graph, N - 1, 1, b, 0);
            return minCostFlow(graph, N - 1, N - 2, a + b)[1];
        }

        public static void addEdge(List<TaskE.Edge>[] graph, int s, int t, int cap, double cost) {
            graph[s].add(new TaskE.Edge(t, cap, cost, graph[t].size()));
            graph[t].add(new TaskE.Edge(s, 0, -cost, graph[s].size() - 1));
        }

        static void bellmanFord(List<TaskE.Edge>[] graph, int s, double[] dist) {
            int n = graph.length;
            Arrays.fill(dist, Integer.MAX_VALUE);
            dist[s] = 0;
            boolean[] inqueue = new boolean[n];
            int[] q = new int[n];
            int qt = 0;
            q[qt++] = s;
            for (int qh = 0; (qh - qt) % n != 0; qh++) {
                int u = q[qh % n];
                inqueue[u] = false;
                for (int i = 0; i < graph[u].size(); i++) {
                    TaskE.Edge e = graph[u].get(i);
                    if (e.cap <= e.f)
                        continue;
                    int v = e.to;
                    double ndist = dist[u] + e.cost;
                    if (dist[v] > ndist + 1e-12) {
                        dist[v] = ndist;
                        if (!inqueue[v]) {
                            inqueue[v] = true;
                            q[qt++ % n] = v;
                        }
                    }
                }
            }
        }

        public static double[] minCostFlow(List<TaskE.Edge>[] graph, int s, int t, int maxf) {
            int n = graph.length;
            double[] prio = new double[n];
            int[] curflow = new int[n];
            int[] prevedge = new int[n];
            int[] prevnode = new int[n];
            double[] pot = new double[n];
            bellmanFord(graph, s, pot);

            int flow = 0;
            double flowCost = 0;
            while (flow < maxf) {
                PriorityQueue<Pair<Double, Integer>> q = new PriorityQueue<>();
                q.add(new Pair<>(0.0, s));
                Arrays.fill(prio, Integer.MAX_VALUE);
                prio[s] = 0;
                boolean[] finished = new boolean[n];
                curflow[s] = Integer.MAX_VALUE;
                while (!finished[t] && !q.isEmpty()) {
                    Pair<Double, Integer> c = q.remove();
                    int u = c.v;
                    double priou = c.u;
                    if (priou != prio[u])
                        continue;
                    finished[u] = true;
                    for (int i = 0; i < graph[u].size(); i++) {
                        TaskE.Edge e = graph[u].get(i);
                        if (e.f >= e.cap)
                            continue;
                        int v = e.to;
                        double nprio = prio[u] + e.cost + pot[u] - pot[v];
                        if (prio[v] > nprio + 1e-12) {
                            prio[v] = nprio;
                            q.add(new Pair<>(nprio, v));
                            prevnode[v] = u;
                            prevedge[v] = i;
                            curflow[v] = Math.min(curflow[u], e.cap - e.f);
                        }
                    }
                }
                if (prio[t] == Integer.MAX_VALUE)
                    break;
                for (int i = 0; i < n; i++)
                    if (finished[i])
                        pot[i] += prio[i] - prio[t];
                int df = Math.min(curflow[t], maxf - flow);
                flow += df;
                for (int v = t; v != s; v = prevnode[v]) {
                    TaskE.Edge e = graph[prevnode[v]].get(prevedge[v]);
                    e.f += df;
                    graph[v].get(e.rev).f -= df;
                    flowCost += df * e.cost;
                }
            }
            return new double[]{flow, flowCost};
        }

        static class Edge {
            int to;
            int f;
            int cap;
            int rev;
            double cost;

            Edge(int to, int cap, double cost, int rev) {
                this.to = to;
                this.cap = cap;
                this.cost = cost;
                this.rev = rev;
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }

    static class Pair<U extends Comparable<U>, V extends Comparable<V>> implements Comparable<Pair<U, V>> {
        public final U u;
        public final V v;

        public Pair(U u, V v) {
            this.u = u;
            this.v = v;
        }

        public int hashCode() {
            return (u == null ? 0 : u.hashCode() * 31) + (v == null ? 0 : v.hashCode());
        }

        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;
            Pair<U, V> p = (Pair<U, V>) o;
            return (u == null ? p.u == null : u.equals(p.u)) && (v == null ? p.v == null : v.equals(p.v));
        }

        public int compareTo(Pair<U, V> b) {
            int cmpU = u.compareTo(b.u);
            return cmpU != 0 ? cmpU : v.compareTo(b.v);
        }

    }
}