Problem:
Score : 500 points

Problem Statement

You are going to hold a competition of one-to-one game called AtCoder Janken. (Janken is the Japanese name for Rock-paper-scissors.)
N players will participate in this competition, and they are given distinct integers from 1 through N.
The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive).
You cannot assign the same integer to multiple playing fields.
The competition consists of N rounds, each of which proceeds as follows:

For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there.

Then, each player adds 1 to its integer. If it becomes N+1, change it to 1.

You want to ensure that no player fights the same opponent more than once during the N rounds.
Print an assignment of integers to the playing fields satisfying this condition.
It can be proved that such an assignment always exists under the constraints given.

Constraints

1 \leq M

M \times 2 +1 \leq N \leq 200000

Input

Input is given from Standard Input in the following format:

N M

Output

Print M lines in the format below.
The i-th line should contain the two integers a_i and b_i assigned to the i-th playing field.

a_1 b_1
a_2 b_2
:
a_M b_M

Sample Input 1

4 1

Sample Output 1

2 3

Let us call the four players A, B, C, and D, and assume that they are initially given the integers 1, 2, 3, and 4, respectively.

The 1-st round is fought by B and C, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 2, 3, 4, and 1, respectively.

The 2-nd round is fought by A and B, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 3, 4, 1, and 2, respectively.

The 3-rd round is fought by D and A, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 4, 1, 2, and 3, respectively.

The 4-th round is fought by C and D, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 1, 2, 3, and 4, respectively.

No player fights the same opponent more than once during the four rounds, so this solution will be accepted.

Sample Input 2

7 3

Sample Output 2

1 6
2 5
3 4

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
 


public class Main {
	private static Scanner sc = new Scanner(System.in);
	//result = Math.pow(num1, num3)
	//StringBuffer str = new StringBuffer(hoge1);
	//String hoge2 = str.reverse().toString();
	//map.containsKey(A)

	//Map<String, Integer> map = new HashMap<String, Integer>(n);
	/*for ( キーのデータ型 key : マップの名前.keySet() ) {
		データのデータ型 data = マップの名前.get( key );
		
		// keyやdataを使った処理;
	}*/
	//int i = Integer.parseInt(s);
	//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek
	//Deque<String> qq=new ArrayDeque<>();//push,pop,peek
	//ArrayList<String> cc = new ArrayList<>(n);
	//Collections.sort(list);
	//Array.sort(list);   cc.contains(tmp)
	//Arrays.asList(c).contains("a")
	//list.set(1,"walk");//1 1 2 3 5
	//static long mod =1000000007;
	//static 	ArrayList<Integer> cc = new ArrayList<>(100);
	//for(int i=0;i<N;i++) {}
	static int a[]=new int[51];
	static int b[]=new int[51];
	static int c[]=new int[51];
	static int d[]=new int[51];
	
	
	
	
	public static void main(String[] args) {
		//String S=sc.next();
		
		//int N=sc.nextInt();for(int i=0;i<N;i++) {}
		int N=sc.nextInt();
		int M=sc.nextInt();
		
		
		int a=N/2;
		int b=a+1;
		
		int x[]=new int[M];
		int y[]=new int[M];
		
		boolean f[]=new boolean[N+1];
		
		
		for(int i=1;i<=M;i++) {
			for(int aa=1;aa+i<=N;aa++) {
				
				if(!f[aa]&&!f[aa+i]) {
					f[aa]=true;
					f[(int) (aa+i)]=true;
					p(aa+" "+(aa+i));
					
					break;
				}
				
			}
			
			
			
		}
		
	
		
		
	
		
		
		
		
	}
	
	
	/*int sum=0;
	 * for(int i=0;i<Q;i++) {
						if(t[b[i]-1]-t[a[i]-1]==c[i]) {
							sum+=d[i];
						}
					}
					ans=Math.max(ans, sum);
					*/
	
	
	
	
	

	


	
	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p() {System.out.println();};
	static void p(long ans) {System.out.println(ans);};
	static void p(double ans) {System.out.println(ans);};
	static String nextPermutation(String s) {
		ArrayList<Character> list=new ArrayList<>();
		for(int i=0;i<s.length();i++) {
			list.add(s.charAt(i));
		}
		int pivotPos=-1;
		char pivot=0;
		for(int i=list.size()-2;i>=0;i--) {
			if(list.get(i)<list.get(i+1)) {
				pivotPos=i;
				pivot=list.get(i);
				break;
			}
		}
		if(pivotPos==-1&&pivot==0) {
			return "Final";
		}
		int L=pivotPos+1,R=list.size()-1;
		
		
		
		int minPos=-1;
		char min =Character.MAX_VALUE;
		
		for(int i=R;i>=L;i--) {
			if(pivot<list.get(i)) {
				if(list.get(i)<min) {
					min=list.get(i);
					minPos=i;
				}
			}
		}
		
		Collections.swap(list, pivotPos, minPos);
		Collections.sort(list.subList(L, R+1));
		
		
		StringBuilder sb=new StringBuilder();
		for(int i=0;i<list.size();i++) {
			sb.append(list.get(i));
		}
		return sb.toString();
	}
}