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.io.*;
import java.util.*;

/**
 *
 * @author Nurgaziyev
 */
public class D2A {

        public void main() throws Exception {
                Scanner in = new Scanner(System.in);

                int n = in.nextInt();
                int k = in.nextInt();

                int[] a = new int[k];
                boolean[] f = new boolean[n * k + 1];
                for (int i = 0; i < k; i++) {
                        a[i] = in.nextInt();
                        f[a[i]] = true;
                }

                for (int i = 0; i < k; i++) {
                        System.out.print(a[i] + " ");
                        int j = 1;
                        while (j < n) {
                                int x = 1;
                                while (f[x]) {
                                        x++;
                                }
                                System.out.print(x + " ");
                                f[x] = true;
                                j++;
                        }
                        System.out.println();
                }
        }

        public static void main(String args[]) throws Exception {
                new D2A().main();
        }

        public class Scanner {

                private BufferedReader br;
                private StringTokenizer st;

                public Scanner(InputStream is) {
                        br = new BufferedReader(new InputStreamReader(is));
                }

                public String next() throws Exception {
                        while (st == null || !st.hasMoreTokens()) {
                                st = new StringTokenizer(br.readLine());
                        }
                        return st.nextToken();
                }

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

                public long nextLong() throws Exception {
                        return Long.parseLong(next());
                }

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