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: Accepted)

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet};
#[allow(unused_imports)]
use std::io::*;
#[allow(unused_imports)]
use std::str::*;

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let cin = stdin();
    let cin = cin.lock();
    let s: String = cin
        .bytes() // Bytes
        .map(|c| c.expect("failed reading char") as char)
        .skip_while(|c| c.is_whitespace()) // c が whitespace である限り skip
        .take_while(|c| !c.is_whitespace()) // c が whitespace でない限り要素を返す
        .collect();
    s.parse().ok().expect("failed parsing")
}

type Ans = Vec<(usize, usize)>;
fn solve(N: usize, M: usize) -> Ans {
    let mut v = Ans::new();
    for i in 0..M {
        let a = i + 1;
        let mut b = i + (M - 1 - i) * 2 + 1 + 1;
        if N % 2 == 0 && b - a >= N / 2 {
            b += 1;
        }
        v.push((a, b));
    }
    v
}

fn test() {
    for M in 1..100 {
        let MIN_N = M * 2 + 1;
        for N in MIN_N..300 {
            dbg!(M, N);
            let ans = solve(N, M);
            let mut h = HashSet::new();
            for i in 0..N {
                for (a, b) in &ans {
                    let mut c = (a + i) % N;
                    let mut d = (b + i) % N;
                    if c > d {
                        let tmp = c;
                        c = d;
                        d = tmp;
                    }
                    if h.contains(&(c, d)) {
                        dbg!((c, d));
                        //dbg!(h);
                        dbg!(i);
                        dbg!(ans);
                        return;
                    }
                    h.insert((c, d));
                }
            }
        }
    }
}

#[allow(non_snake_case)]
fn main() {
    input! {
        N:usize,
        M:usize
    }
    let ans = solve(N, M);
    for (a, b) in &ans {
        println!("{} {}", a, b);
    }
}