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 taskB {
	FastScanner in;
	PrintWriter out;

	public void solve() throws IOException {
		int n=in.nextInt();
		int k=in.nextInt();
		int use[]= new int [n*k+1];
		int a[]=new int[k+1];
		Arrays.fill(use, -1);
		for (int i=0;i<k;i++){
			int x=in.nextInt();
			use[x]=i+1;
			a[i+1]++;
		}
		
		for (int i=1;i<=k;i++)
				for (int j=1;j<=n*k && a[i] <n;j++)
					if (use[j]==-1){
						use[j]=i;
						a[i]++;
					}
		
		for (int i=1;i<=k;i++){
			for (int j=1;j<=n*k;j++)
				if (use[j]==i){
					out.print(j+" ");
				}
			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(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				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());
		}
	}

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