Problem: Vasya is about to take his first university exam in about several minutes. And it's not just some ordinary exam, it's on mathematical analysis. Of course, right now Vasya can only think of one thing: what the result of his talk with the examiner will be...

To prepare for the exam, one has to study proofs of n theorems. It is known that there will be k examination cards on the exam and each card contains <image> distinct theorems. Besides, no theorem is mentioned in more than one card (that is, <image> theorems won't be mentioned in any card). During the exam several students may get the same card.

We do not know the exact way theorems are distributed by cards, however the students that took the exam before Vasya told him what theorems their cards contained. Vasya evaluates his level of proficiency in the i-th theorem by some number ai. The level of proficiency in some card is the average of the levels of proficiency in the theorems that are included in the card. Now Vasya wants to know the minimally and maximally possible levels of his proficiency in the card he gets on the exam. Vasya wants to determine it by the data he has collected from other students. Unfortunately, Vasya has no time left to do the math and he asked you to help him.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 100) — the number of theorems and the number of cards correspondingly. The second line contains n integers ai (0 ≤ ai ≤ 100), the i-th number (1 ≤ i ≤ n) corresponds to Vasya's proficiency in the i-th theorem.

The third line contains number q (0 ≤ q ≤ 100) — the number of people that have taken the exam before Vasya. Each of the following q lines contains the description of a student's card: <image> integers from 1 to n inclusive. They are the numbers of theorems included in the card in the order in which they are enumerated in the input data. The numbers are given in an arbitrary order. It is guaranteed that the given cards are valid (that is, that all theorems in one card are different and that different people get cards that either don't contain the same theorems or coincide up to the theorems' permutation).

Output

Print two real numbers, representing Vasya's minimum and maximum proficiency in the card he will get on the exam. The absolute or relative error should not exceed 10 - 6.

Examples

Input

7 3
7 15 0 19 10 5 12
2
1 6
7 4


Output

5.0000000000 15.5000000000

Input

4 2
10 8 1 17
2
2 3
3 2


Output

4.5000000000 13.5000000000

Note

Let's analyze the first sample. Vasya's proficiency in the cards whose content he already knows equals 6 and 15.5 correspondingly. The three theorems that are left are only enough to make one exam card. If we consider all possible variants of theorems included in the card we can see that in the best case scenario Vasya gets the card that contains theorems 4 and 7 (his proficiency would equal 15.5) and in the worst case scenario he gets theorems 3 and 5 (his proficiency would equal 5).

The ⌊ x⌋ operation denotes taking integer part of real number x (rounding down).

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class a {
    static long mod = 1000000009;
    static ArrayList<Integer>[] tree;
    static long[] data;
public static void main(String[] args) throws IOException
{
    //Scanner input = new Scanner(new File("input.txt"));
    //PrintWriter out = new PrintWriter(new File("output.txt"));
  input.init(System.in);
  PrintWriter out  = new PrintWriter((System.out));
  int n = input.nextInt(), m = input.nextInt();
  double count = n/m;
  int[] data = new int[n];
  for(int i = 0; i<n; i++) data[i] = input.nextInt();
  boolean[] used = new boolean[n];
  double min = 200, max = 0;
  int q = input.nextInt();
  for(int i = 0;i<q; i++)
  {
      double sum = 0;
      for(int j = 0; j<count; j++)
      {
          int x = input.nextInt()-1;
          used[x] = true;
          sum += data[x];
      }
      sum/=count;
      min = Math.min(min, sum);
      max = Math.max(max, sum);
  }
  ArrayList<Integer> left = new ArrayList<Integer>();
  for(int i = 0; i<n;i++)
      if(!used[i])
          left.add(data[i]);
  if(left.size()>=count && (n-left.size())/count < m)
  {
      Collections.sort(left);
      double minsum = 0, maxsum = 0;
      for(int i = 0; i<count; i++)
      {
          minsum += left.get(i);
          maxsum += left.get(left.size()-i-1);
      }
      minsum /= count;
      maxsum /= count;
      min = Math.min(min, minsum);
      max = Math.max(max, maxsum);
  }
  out.println(min+" "+max);
  out.close();
}
static long pow(long x, long p)
{
    if(p==0) return 1;
    if((p&1) > 0)
    {
        return (x*pow(x, p-1))%mod;
    }
    long sqrt = pow(x, p/2);
    return (sqrt*sqrt)%mod;
}
static long gcd(long a, long b)
{
    if(b==0) return a;
    return gcd(b, a%b);
}
static class input {
  static BufferedReader reader;
  static StringTokenizer tokenizer;

  /** call this method to initialize reader for InputStream */
  static void init(InputStream input) {
      reader = new BufferedReader(
                   new InputStreamReader(input) );
      tokenizer = new StringTokenizer("");
  }

  /** get next word */
  static String next() throws IOException {
      while ( ! tokenizer.hasMoreTokens() ) {
          //TODO add check for eof if necessary
          tokenizer = new StringTokenizer(
                 reader.readLine() );
      }
      return tokenizer.nextToken();
  }

  static int nextInt() throws IOException {
      return Integer.parseInt( next() );
  }
  
  static double nextDouble() throws IOException {
      return Double.parseDouble( next() );
  }
  static long nextLong() throws IOException {
      return Long.parseLong( next() );
  }
  static String nextLine() throws IOException {
    return reader.readLine();
  }
}
/*
Sum Interval Tree - uses O(n) space
Updates and queries over a range of values in log(n) time
Note: If querying all elements, use difference array instead.
*/
static class IT
{
    int[] left,right, val, a, b;
    IT(int n)
    {
        left = new int[4*n];
        right = new int[4*n];
        val = new int[4*n];
        a = new int[4*n];
        b = new int[4*n];
        init(0,0, n);
    }
    int init(int at, int l, int r)
    {
        a[at] = l;
        b[at] = r;
        if(l==r)
            left[at] = right [at] = -1;
        else 
        {
            int mid = (l+r)/2;
            left[at] = init(2*at+1,l,mid);
            right[at] = init(2*at+2,mid+1,r);
        }
        return at++;
    }
    //return the sum over [x,y]
    int get(int x, int y)
    {
        return go(x,y, 0);
    }
    int go(int x,int y, int at)
    {
        if(at==-1) return 0;
        if(x <= a[at] && y>= b[at]) return val[at];
        if(y<a[at] || x>b[at]) return 0;
        return go(x, y, left[at]) + go(x, y, right[at]);
    }
    //add v to elements x through y
    void add(int x, int y, int v)
    {
        go3(x, y, v, 0);
    }
    void go3(int x, int y, int v, int at)
    {
        if(at==-1) return;
        if(y < a[at] || x > b[at]) return;
        val[at] += (Math.min(b[at], y) - Math.max(a[at], x) + 1)*v;
        go3(x, y, v, left[at]);
        go3(x, y, v, right[at]);
    }
}
}