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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringJoiner;
import java.util.StringTokenizer;
import java.util.function.IntUnaryOperator;
import java.util.function.LongUnaryOperator;

public class Main {
    static In in = new In();
    static Out out = new Out();
    static final long mod = 1000000007;
    static final long inf = 0xfffffffffffffffL;
    static final int iinf = 0xfffffff;
    static final int[] da = {-1, 0, 1, 0, -1, 1, 1, -1, -1};

    int rem;
    void solve() {
        int n = in.nextInt();
        int m = in.nextInt();
        rem = m;
        f(n, m);
        g(n, m);
    }

    void f(int n, int m) {
        int h = (n + 1) / 2 / 2 * 2;
        int i = 0;
        while (i < h) {
            println(i + 1, h);
            h--;
            i++;
        }
    }

    void g(int n, int m) {
        int i = (n + 1) / 2 / 2 * 2 + 1;
        int j = n;
        while (i < j) {
            println(i, j);
            i++;
            j--;
        }
    }

    void println(int a, int b) {
        out.println(a, b);
        rem--;
        if (rem == 0) {
            out.flush();
            System.exit(0);
        }
    }


    public static void main(String[]$) {
        new Main().solve();
        out.flush();
    }
}

class In {
    private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 0x10000);
    private StringTokenizer tokenizer;

    String next() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
        } catch (IOException ignored) {
        }
        return tokenizer.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    char[][] nextCharArray(int n, int m) {
        char[][] a = new char[n][m];
        for (int i = 0; i < n; i++) {
            a[i] = next().toCharArray();
        }
        return a;
    }

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

    int[] nextIntArray(int n, IntUnaryOperator op) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsInt(nextInt());
        }
        return a;
    }

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

    long[] nextLongArray(int n, LongUnaryOperator op) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsLong(nextLong());
        }
        return a;
    }
}

class Out {
    private PrintWriter out = new PrintWriter(System.out);
    boolean autoFlush = false;

    void println(Object... a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (Object obj : a) {
            joiner.add(String.valueOf(obj));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void println(int[] a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (int i : a) {
            joiner.add(Integer.toString(i));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void println(long[] a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (long i : a) {
            joiner.add(Long.toString(i));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void flush() {
        out.flush();
    }
}