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

public class A {
    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int[] ans = new int[n * k];
        for (int i = 0; i < k; i++) {
            ans[in.nextInt() - 1] = i + 1;
        }
        int cur = n - 1;
        int curIdx = 1;
        for (int i = 0; i <  k * n; i++) {
            if (ans[i] == 0) {
                ans[i] = curIdx;
                cur--;
            }
            if (cur == 0) {
                cur = n - 1;
                curIdx++;
            }
        }
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j < n * k; j++) {
                if (ans[j] == i) {
                    out.print((j + 1) + " ");
                }
            }
            out.println();
        }
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
        }

        String next() throws IOException {
            return hasNext() ? st.nextToken() : null;
        }

        boolean hasNext() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                String line = br.readLine();
                if (line == null) {
                    return false;
                }
                st = new StringTokenizer(line);
            }
            return true;
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.parseInt(next());
        }
    }

    public static void main(String[] arg) {
        new A().run();
    }
}