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.*;

public class AR150 {
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;

	void solve() throws IOException {
		int n = nextInt();
		int k = nextInt();
		int[] a = new int[k + 1];
		boolean[] used = new boolean[n * k + 1];
		Arrays.fill(used, false);
		for (int i = 0; i < k; i++){
			a[i] = nextInt();
			used[a[i]] = true;
		}
		int j = 1;
		for (int i = 0; i < k; i++){
			out.print(a[i] + " ");
			int l = 1;
			while (l != n){
				if (!used[j]){
					out.print(j + " ");
					l++;
				}
				j++;
			}
			out.println();
		}
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(new OutputStreamWriter(System.out));

		solve();

		in.close();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new AR150().run();
	}

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

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

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

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

}