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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main {


// builder
public static StringBuilder br;

public static void add(String cad) {
	if (br == null)
		br = new StringBuilder();
	br.append(cad).append("\n");
}

// split de array
public static int[] atoi(String cad) {
	String read[] = cad.split(" ");
	int res[] = new int[read.length+1];
	for (int i = 0; i < read.length; i++) {
		res[i+1] = Integer.parseInt(read[i]);
	}
	return res;
}

// split de String
public static String[] atos(String cad) {
	return cad.split(" ");
}

// parsing de String a int
public static int parseo(String cad, int index) {
	return Integer.parseInt(cad.split(" ")[index]);
}

// memory status
static String memoryStatus() {
	return (Runtime.getRuntime().totalMemory()
			- Runtime.getRuntime().freeMemory() >> 20)
			+ "/" + (Runtime.getRuntime().totalMemory() >> 20) + " MB";
}

// imprimir array
static void printArrayInt(int[] array) {
	if (array == null || array.length == 0)
		return;
	for (int i = 0; i < array.length; i++) {
		if (i > 0)
			System.out.print(" ");
		System.out.print(array[i]);
	}
	System.out.println();
}

// impresion de datos
static void print(Object... obj) {
	for (int i = 0; i < obj.length; i++) {
		if (i != 0)
			System.out.print(" ");
		System.out.print(obj[i]);
	}
	System.out.println();
}

public static void main(String[] args) throws IOException {
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
			String line = in.readLine();;
			int n=parseo(line,0);
			int p=parseo(line,1);
			int arr[]=atoi(in.readLine());
			int cad[]=new int[(n*p)+1];
			for (int i = 0; i < arr.length; i++) {
				cad[arr[i]]=-1;
			}
			for (int i = 1; i <= p; i++) {
				if(n!=1)System.out.print(arr[i]+" ");
				else System.out.println(arr[i]);
				int l=n*p;
				int cont=1;
				ciclo1:for (int j = 1; j <=l ; j++) {
					while( j!=l && cad[j]!=0){
						j++;
					}
					if(cad[j]==0){
						if(cont!=n-1)System.out.print(j+" ");
						else System.out.print(j);
						cad[j]=-1;
						cont++;
						if(cont==n)break ciclo1;;
					}			
				}
				System.out.println();
			}
			
	
	//System.out.print(br != null ? br : "");
}
}