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(dead_code)]
fn main() {
    let (n, m) = i::u2();
    let even = (m + 1) / 2;
    let odd = m - even;
    if m % 2 == 0 {
        for i in 0..even {
            println!("{} {}", i + 1, m - i + 1);
        }
        for i in 0..odd {
            println!("{} {}", m + i + 2, m + odd * 2 - i + 1);
        }
    } else {
        for i in 0..even {
            println!("{} {}", i + 1, m - i + 1);
        }
        for i in 0..odd {
            println!("{} {}", m + i + 2, m + odd * 2 - i + 2);
        }
    }
}

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

#[allow(dead_code)]
mod i {
    use super::*;

    pub fn read<T: FromStr>() -> T {
        stdin()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>()
            .parse::<T>()
            .ok()
            .unwrap()
    }

    pub fn str() -> String {
        read()
    }

    pub fn s() -> Vec<char> {
        str().chars().collect()
    }

    pub fn i() -> i64 {
        read()
    }

    pub fn u() -> usize {
        read()
    }

    pub fn u2() -> (usize, usize) {
        (read(), read())
    }

    pub fn u3() -> (usize, usize, usize) {
        (read(), read(), read())
    }

    pub fn f() -> f64 {
        read()
    }

    pub fn c() -> char {
        read::<String>().pop().unwrap()
    }

    pub fn iv(n: usize) -> Vec<i64> {
        (0..n).map(|_| i()).collect()
    }

    pub fn iv2(n: usize) -> Vec<(i64, i64)> {
        (0..n).map(|_| iv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv(n: usize) -> Vec<usize> {
        (0..n).map(|_| u()).collect()
    }

    pub fn uv2(n: usize) -> Vec<(usize, usize)> {
        (0..n).map(|_| uv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv3(n: usize) -> Vec<(usize, usize, usize)> {
        (0..n).map(|_| uv(3)).map(|a| (a[0], a[1], a[2])).collect()
    }

    pub fn fv(n: usize) -> Vec<f64> {
        (0..n).map(|_| f()).collect()
    }

    pub fn cmap(h: usize) -> Vec<Vec<char>> {
        (0..h).map(|_| s()).collect()
    }
}