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 Rust (1.42.0) (judge verdict: Wrong Answer)

//! Framework <https://github.com/vain0x/procon>

#![allow(non_snake_case)]
#![allow(unused_imports)]

use std::collections::*;

pub struct Scan(Box<dyn Iterator<Item = &'static str>>); // '

impl Scan {
    fn new() -> Self {
        let mut buf = String::new();
        let read_line = move || {
            std::io::stdin().read_line(&mut buf).unwrap();
            Box::leak(buf.split_off(0).into_boxed_str()).split_whitespace()
        };
        Scan(Box::new(std::iter::repeat_with(read_line).flatten()))
    }

    pub fn word<T: std::str::FromStr>(&mut self) -> T {
        self.0.next().unwrap().parse().ok().unwrap()
    }

    pub fn list<T: std::str::FromStr>(&mut self, len: usize) -> Vec<T> {
        std::iter::repeat_with(|| self.word()).take(len).collect()
    }
}

fn validate(N: usize, M: usize, assignment: &[(usize, usize)]) -> bool {
    let mut freq = HashMap::new();

    for x in 0..N {
        for i in 0..M {
            let (u, v) = assignment[i];

            let u = (u + x - 1) % N + 1;
            let v = (v + x - 1) % N + 1;
            *freq.entry((u, v)).or_insert(0) += 1_usize;
            *freq.entry((v, u)).or_insert(0) += 1_usize;

            if freq[&(u, v)] > 1 || freq[&(v, u)] > 1 {
                #[cfg(debug_assertions)]
                eprintln!(
                    "\x1B[33m{}\x1B[0m = {:?}",
                    "(N, M, &assignment, x, i, u, v)",
                    (N, M, &assignment, x, i, u, v)
                );
                return false;
            }
        }
    }

    let mut freq = HashMap::new();
    for (u, v) in assignment {
        *freq.entry(u).or_insert(0) += 1_usize;
        *freq.entry(v).or_insert(0) += 1_usize;
    }

    for u in 1..=N {
        if freq.get(&u).cloned().unwrap_or(0) > 1 {
            #[cfg(debug_assertions)]
            eprintln!("\x1B[33m{}\x1B[0m = {:?}", "u", u);
            return false;
        }
    }

    true
}

fn check() {
    let mut assignment = vec![];
    for N in 3..30 {
        let mut M = 1;
        while M * 2 + 1 <= N {
            assignment.clear();
            for x in 0..M {
                assignment.push((x + 1, x + M - 1 + (x + 1) + 1));
            }

            if !validate(N, M, &assignment) {
                break;
            }

            M += 1;
        }
    }
}

fn main() {
    let mut scan = Scan::new();

    let N = scan.word::<usize>();
    let M = scan.word::<usize>();

    let mut assignment = vec![];
    let mut done = vec![false; N + 1];
    let mut l = 0;
    let mut r = N;

    for x in (1..=M).rev() {
        let (mut i, step) = if x % 2 == 0 { (l, 1) } else { (r - x - 1, -1) };

        while l <= i && i + x < r {
            if !done[i] && !done[i + x] {
                assignment.push((i + 1, i + x + 1));

                done[i] = true;
                if i == l {
                    l += 1;
                }

                done[i + x] = true;
                if i + x == r - 1 {
                    r -= 1;
                }
                break;
            }

            i = (i as isize + step) as usize;
        }
    }

    debug_assert!(validate(N, M, &assignment));

    for &(u, v) in &assignment {
        println!("{} {}", u + 1, v + 1);
    }
}