Problem: One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k. 

There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i-th (1 ≤ i ≤ k) child wrote the number ai (1 ≤ ai ≤ n·k). All numbers ai accidentally turned out to be different.

Now the children wonder, how to divide the orange so as to meet these conditions:

  * each child gets exactly n orange segments; 
  * the i-th child gets the segment with number ai for sure; 
  * no segment goes to two children simultaneously. 



Help the children, divide the orange and fulfill the requirements, described above.

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains k space-separated integers a1, a2, ..., ak (1 ≤ ai ≤ n·k), where ai is the number of the orange segment that the i-th child would like to get.

It is guaranteed that all numbers ai are distinct.

Output

Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.

You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.

Examples

Input

2 2
4 1


Output

2 4 
1 3 


Input

3 1
2


Output

3 2 1 

Solution (in JAVA):

Solution:
import java.awt.Point;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.TreeMap;

public class A{

    
    static Point end;

    public static void main(String[] Args) {
        int n=si(), k=si();
        
        int[] a = new int[k+1];
        for (int i = 1; i <=k; i++) {
            a[i]=si();
        }
        
        ArrayList<Integer> segments = new ArrayList<Integer>();
        for (int i = 1; i <= n*k; i++) {
            segments.add(i);
        }
        TreeMap<Integer, ArrayList<Integer>> childs = new TreeMap<Integer, ArrayList<Integer>>();
        for (int i = 1; i <=k; i++) {
            childs.put(i, new ArrayList<Integer>());
        }
        
        //remove segment wanted
        for (int i = 1; i <=k; i++) {
            childs.get(i).add(a[i]);
            segments.remove(new Integer(a[i]));
        }
        
        //distribute
        while(segments.size()>0){
            for(Integer c: childs.keySet()){
                childs.get(c).add(segments.remove(0));
            }
        }
        for(int c: childs.keySet()){
            for(int i: childs.get(c)){
                System.out.print(i+" ");
            }
            System.out.println("");
        }
    }

    static int toi(Object s) {
        return Integer.parseInt(s.toString());
    }

    // ----------------------- Library ------------------------
    static int[] dx = { 1, 0, -1, 1, -1, 1, 0, -1 }, dy = { 1, 1, 1, 0, 0, -1,
            -1, -1 };
    static Scanner scan = new Scanner(System.in);
    static int INF = 2147483647;

    // finds GCD of a and b using Euclidian algorithm
    public int GCD(int a, int b) {
        if (b == 0)
            return a;
        return GCD(b, a % b);
    }

    static List<String> toList(String[] a) {
        return Arrays.asList(a);
    }

    static String[] toArray(List<String> a) {
        String[] o = new String[a.size()];
        a.toArray(o);
        return o;
    }

    static int[] pair(int... a) {
        return a;
    }

    static int si() {
        return scan.nextInt();
    }

    static String ss() {
        return scan.nextLine();
    }

    static int[] sai(int n) {
        int[] a = new int[n];
        for (int i = 0; i < a.length; i++)
            a[i] = si();

        return a;
    }

    static String[] sas(int n) {
        String[] a = new String[n];
        for (int i = 0; i < a.length; i++)
            a[i] = ss();
        return a;
    }

    static Object[][] _sm1(int r, int c) {
        Object[][] a = new Object[r][c];
        for (int i = 0; i < r; i++)
            for (int j = 0; j < c; j++)
                a[i][j] = scan.next();
        return a;
    }

    static Object[][] _sm2(int r) {
        Object[][] a = new Object[r][3];
        for (int i = 0; i < r; i++)
            a[i] = new Object[] { ss(), ss(), ss() };
        return a;
    }
}