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.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class Main {
  public static void main(String[] args) {
    MyScanner sc = new MyScanner();
    int n = sc.nextInt();   // number of ints in the array
    int k = sc.nextInt();   // number of parts
    int ke = sc.nextInt();  // number of parts with even sum
    int ko = k - ke;
    int numOdd = 0;         // number of odd entries
    int[] arr = new int[n];
    for (int i = 0; i < n; ++i) {
      arr[i] = sc.nextInt();
      if (arr[i] % 2 == 1) {
        numOdd++;
      }
    }
    int numEven = n - numOdd;
    
    // Make sure that the number of odd numbers has the correct parity.
    if (numOdd % 2 != ko % 2) {
      System.out.println("NO");
      return;
    }
    
    // Make sure that we have enough odd numbers.
    if (numOdd < ko) {
      System.out.println("NO");
      return;
    }
    
    // At this point we know that numOdd >= ko.  Exactly 'ko' of the odd numbers will be chosen to
    // be its own part.  Make sure that we have enough leftover numbers to fill in the even parts.
    // Note that (numOdd - ko) is guaranteed to be a multiple of two.
    int availableEvenParts = (numOdd - ko) / 2 + numEven;
    if (availableEvenParts < ke) {
      System.out.println("NO");
      return;
    }
    
    int[] odds = new int[numOdd];
    int[] evens = new int[numEven];
    int io = 0;
    int ie = 0;
    for (int i = 0; i < n; ++i) {
      if (arr[i] % 2 == 0) {
        evens[ie++] = arr[i];
      } else {
        odds[io++] = arr[i];
      }
    }
     
    System.out.println("YES");
    io = 0;
    ie = 0;
    // Print each odd number in its own part, except for possibly the last part, which has the rest
    // of the integers.
    if (ke == 0) {
      for (int i = 0; i < ko - 1; ++i) {
        System.out.println("1 " + odds[io++]);
      }
      printFinalPart(numOdd, io, numEven, ie, odds, evens);
      return;
    }
    // Print each odd number in its own part, and then print the remaining with exactly two odds or
    // one even in each part, except for possibly the last part.
    for (int i = 0; i < ko; ++i) {
      System.out.println("1 " + odds[io++]);
    }
    for (; ke != 0; --ke) {
      if (ke != 1) {
        if (io != numOdd) {
          System.out.println("2 " + odds[io] + " " + odds[io + 1]);
          io += 2;
        } else {
          System.out.println("1 " + evens[ie++]);
        }
      } else {
        printFinalPart(numOdd, io, numEven, ie, odds, evens);
      }
    }
  }
  
  public static void printFinalPart(int numOdd, int io, int numEven, int ie, int[] odds,
      int[] evens) {
    // Print the size of this part
    System.out.print(numOdd - io + numEven - ie + " ");
    // Print all remaining ints.
    while (io != numOdd) {
      System.out.print(odds[io++] + " ");
    }
    while (ie != numEven) {
      System.out.print(evens[ie++] + " ");
    }
  }
  
      public static class MyScanner {
    BufferedReader br;
    StringTokenizer st;
    
    public MyScanner() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }
 
    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }
 
    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }
 
    double nextDouble() {
      return Double.parseDouble(next());
    }
 
    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }
  }
}