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.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.net.ConnectException;
import java.rmi.dgc.Lease;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.Objects;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.util.Comparator.*;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
		main.out.close();
	}

	// ======================================================================
	public void solve() {
		int N = ni();
		int M = ni();
		int start = 1, end = (N/2*2);
		for (int i = 0; i < M; i++) {
			out.println(start + " " + end);
			start++;
			end--;
		}
	}

	// ------------------------------------------
	// ライブラリ
	// ------------------------------------------
	// Print
	private PrintWriter out = new PrintWriter(System.out);

	// Scanner
	private FastScanner scan = new FastScanner();

	int ni() {
		return scan.nextInt();
	}

	int[] ni(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = ni();
		}
		return a;
	}

	int[][] ni(int y, int x) {
		int[][] a = new int[y][x];
		for (int i = 0; i < y; i++) {
			for (int j = 0; j < x; j++) {
				a[i][j] = ni();
			}
		}
		return a;
	}

	long nl() {
		return scan.nextLong();
	}

	long[] nl(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = nl();
		}
		return a;
	}

	long[][] nl(int y, int x) {
		long[][] a = new long[y][x];
		for (int i = 0; i < y; i++) {
			for (int j = 0; j < x; j++) {
				a[i][j] = nl();
			}
		}
		return a;
	}

	String ns() {
		return scan.next();
	}

	String[] ns(int n) {
		String[] a = new String[n];
		for (int i = 0; i < n; i++) {
			a[i] = ns();
		}
		return a;
	}

	String[][] ns(int y, int x) {
		String[][] a = new String[y][x];
		for (int i = 0; i < y; i++) {
			for (int j = 0; j < x; j++) {
				a[i][j] = ns();
			}
		}
		return a;
	}
}

class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}