Problem: Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?

Given an array consisting of distinct integers. Is it possible to partition the whole array into k disjoint non-empty parts such that p of the parts have even sum (each of them must have even sum) and remaining k - p have odd sum? (note that parts need not to be continuous).

If it is possible to partition the array, also give any possible way of valid partitioning.

Input

The first line will contain three space separated integers n, k, p (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will contain n space-separated distinct integers representing the content of array a: a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).

If the required partition exists, print k lines after the first line. The ith of them should contain the content of the ith part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly p parts with even sum, each of the remaining k - p parts must have odd sum.

As there can be multiple partitions, you are allowed to print any valid partition.

Examples

Input

5 5 3
2 6 10 5 9


Output

YES
1 9
1 5
1 10
1 6
1 2


Input

5 5 3
7 14 2 9 5


Output

NO


Input

5 3 1
1 2 3 7 5


Output

YES
3 5 1 3
1 7
1 2

Solution (in JAVA):

Solution:
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.TreeMap;
import java.util.HashSet;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

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

class TaskC {

	public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int k = in.nextInt(); // k partitions
        int p = in.nextInt(); // are even sums
        int q = k - p;  // odd sums
        ArrayList<Integer> odds = new ArrayList<Integer>();
        ArrayList<Integer> evens = new ArrayList<Integer>();
        for (int i = 0; i < n; ++i) {
            int num = in.nextInt();
            if (num % 2 == 1) odds.add(num);
            else evens.add(num);
        }

        if (odds.size() < q) {
            out.write("NO\n");
            return;
        }
        int leftovers = odds.size() - q;
        if (leftovers < 0 || leftovers % 2 == 1 || (leftovers/2 + evens.size()) < p) {
            out.write("NO\n");
            return;
        }

        out.write("YES\n");
        
        int e = 0;
        int o = 0;
        int i = 0;
        // Write odds
        for (; i < q; ++i) {
            if (i == q - 1 && p == 0) break;
            out.write("1 " + odds.get(o) + "\n");
            o++;
        }

        // Write single evens 
        for (i = 0; i < p - 1 && e < evens.size(); ++i) {
            out.write("1 " + evens.get(e) + "\n");
            e++;
        }
        // If there are no odds and one even left, fill it with even
        // If there are no odds and many evens left, fill it with the rest
        // If there are odds and we don't have enough evens, fill it with odds

        // Write pairs of odds until the second to last
        while (i < p - 1) {
            out.write("2 " + odds.get(o) + " " + odds.get(o+1) + "\n");
            o += 2;
            ++i;
        }

        // How many are there
        if (e + o == n) return;

        out.write((n - e - o) + " ");
        // Fill with the rest of the evens
        while (e < evens.size()) {
            out.write(evens.get(e) + " ");
            e++;
        }
        while (o < odds.size()) {
            out.write(odds.get(o) + " ");
            o++;
        }

        
    }

}

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 int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }

}