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:
//Date: Oct 12, 2013 
//Time: 9:58:39 AM

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

public class A244 implements Runnable {

	public void solve() throws IOException {
		Set<Integer> set = new TreeSet<Integer>();
		int N = nextInt();
		int K = nextInt();
		int[] k = new int[K];
		for(int i = 0; i < K; i++){ k[i] = nextInt(); set.add(k[i]); }
		
		int counter = 1;
		for(int i = 0; i < K; i++){
			System.out.print(k[i] + " ");
			for(int j = 0; j < N - 1; j++){
				while(set.contains(counter)) counter++;
				System.out.print(counter + " ");
				counter++;
			}
			System.out.println();
		}
		
	}

	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new A244().run();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

	public String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	BufferedReader in;
	StringTokenizer tok;
}
/**
public class A244 {

}
 
*/