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)

#[derive(Default)]
//NOTE
//declare variables to reduce the number of parameters for dp and dfs etc.
struct Solver {}

impl Solver {
    fn solve(&mut self) {
        let stdin = std::io::stdin();
        let mut scn = Scanner {
            stdin: stdin.lock(),
        };
        let n: usize = scn.read();
        let m: usize = scn.read();

        if (n - 2) % 4 == 0 {

            let mut left = 1;
            let mut right = n;
            let k = (n - 2) / 4;
            for i in 0..m {
                if i == k {
                    left += 1;
                }
                assert!(left < right);
                println!("{} {}", left, right);
                left += 1;
                right -= 1;
            }
        } else {
            let mut left = 1;
            let mut right = n;
            for i in 0..m {
                assert!(left < right);
                println!("{} {}", left, right);
                left += 1;
                right -= 1;
            }
        }
    }
}

fn main() {
    std::thread::Builder::new()
        .stack_size(64 * 1024 * 1024) // 64MB
        .spawn(|| {
            let mut solver = Solver::default();
            solver.solve()
        })
        .unwrap()
        .join()
        .unwrap();
}
pub mod utils {
    const DYX: [(isize, isize); 8] = [
        (0, 1),   //right
        (1, 0),   //down
        (0, -1),  //left
        (-1, 0),  //top
        (1, 1),   //down right
        (-1, 1),  //top right
        (1, -1),  //down left
        (-1, -1), //top left
    ];

    pub fn try_adj(y: usize, x: usize, dir: usize, h: usize, w: usize) -> Option<(usize, usize)> {
        let ny = y as isize + DYX[dir].0;
        let nx = x as isize + DYX[dir].1;
        if ny >= 0 && nx >= 0 {
            let ny = ny as usize;
            let nx = nx as usize;
            if ny < h && nx < w {
                Some((ny, nx))
            } else {
                None
            }
        } else {
            None
        }
    }
}

//snippet from kenkoooo
pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_line(&mut self) -> String {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b'\n' || b == b'\r')
            .take_while(|&b| b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}