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)

#[macro_use]
mod input_mcr {
    // ref: tanakh <https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8>
    // diff: using Parser
    #[macro_export(local_inner_macros)]
    macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut parser = Parser::from_str($s);
        input_inner!{parser, $($r)*}
    };
    (parser = $parser:ident, $($r:tt)*) => {
        input_inner!{$parser, $($r)*}
    };
    (new_stdin_parser = $parser:ident, $($r:tt)*) => {
        let stdin = std::io::stdin();
        let reader = std::io::BufReader::new(stdin.lock());
        let mut $parser = Parser::new(reader);
        input_inner!{$parser, $($r)*}
    };
    ($($r:tt)*) => {
        input!{new_stdin_parser = parser, $($r)*}
    };
}

    #[macro_export(local_inner_macros)]
    macro_rules! input_inner {
    ($parser:ident) => {};
    ($parser:ident, ) => {};
    ($parser:ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($parser, $t);
        input_inner!{$parser $($r)*}
    };
}

    #[macro_export(local_inner_macros)]
    macro_rules! read_value {
    ($parser:ident, ( $($t:tt),* )) => {
        ( $(read_value!($parser, $t)),* )
    };
    ($parser:ident, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($parser, $t)).collect::<Vec<_>>()
    };
    ($parser:ident, chars) => {
        read_value!($parser, String).chars().collect::<Vec<char>>()
    };
    ($parser:ident, char_) => {
        read_value!($parser, String).chars().collect::<Vec<char>>()[0]
    };
    ($parser:ident, usize1) => {
        read_value!($parser, usize) - 1
    };
    ($parser:ident, line) => {
        $parser.next_line()
    };
    ($parser:ident, line_) => {
        $parser.next_line().chars().collect::<Vec<char>>()
    };
    ($parser:ident, i64_) => {
        $parser.fast_i64()
    };
    ($parser:ident, usize_) => {
        $parser.fast_i64() as usize
    };
    ($parser:ident, usize1_) => {
        ($parser.fast_i64() - 1) as usize
    };
    ($parser:ident, $t:ty) => {
        $parser.next::<$t>().expect("Parse error")
    };
}

    use std::io;
    use std::io::BufRead;
    use std::str;

    // ref: tatsuya6502 <https://qiita.com/tatsuya6502/items/cd448486f7ef7b5b8c7e>
    // ref: wariuni <https://qiita.com/tatsuya6502/items/cd448486f7ef7b5b8c7e#comment-7040a5ae96305e884eb9>
    // diff: using std::io::BufRead::fill_buf()
    pub struct Parser<R> {
        pub reader: R,
        buf: Vec<u8>,
        pos: usize,
    }

    impl Parser<io::Empty> {
        pub fn from_str(s: &str) -> Parser<io::Empty> {
            Parser {
                reader: io::empty(),
                buf: s.as_bytes().to_vec(),
                pos: 0,
            }
        }
    }

    impl<R: BufRead> Parser<R> {
        pub fn new(reader: R) -> Parser<R> {
            Parser {
                reader: reader,
                buf: vec![],
                pos: 0,
            }
        }
        pub fn update_buf(&mut self) {
            self.buf.clear();
            self.pos = 0;
            loop {
                let (len, complete) = {
                    let buf2 = self.reader.fill_buf().unwrap();
                    self.buf.extend_from_slice(buf2);
                    let len = buf2.len();
                    (len, buf2[len - 1] <= 0x20)
                };
                self.reader.consume(len);
                if complete {
                    break;
                }
            }
        }
        pub fn next<T: str::FromStr>(&mut self) -> Result<T, T::Err> {
            loop {
                let mut begin = self.pos;
                while begin < self.buf.len() && (self.buf[begin] <= 0x20) {
                    begin += 1;
                }
                let mut end = begin;
                while end < self.buf.len() && (self.buf[end] > 0x20) {
                    end += 1;
                }
                if begin != self.buf.len() {
                    self.pos = end;
                    return unsafe { str::from_utf8_unchecked(&self.buf[begin..end]) }.parse::<T>();
                } else {
                    self.update_buf();
                }
            }
        }
        pub fn next_line(&mut self) -> String {
            loop {
                let mut begin = self.pos;
                if begin < self.buf.len() && (self.buf[begin] < 0x20) {
                    begin += 1;
                }
                let mut end = begin;
                while end < self.buf.len() && (self.buf[end] >= 0x20) {
                    end += 1;
                }
                if begin != self.buf.len() {
                    self.pos = end;
                    return unsafe { str::from_utf8_unchecked(&self.buf[begin..end]) }.to_string();
                } else {
                    self.update_buf();
                }
            }
        }
        pub fn fast_i64(&mut self) -> i64 {
            loop {
                let mut begin = self.pos;
                while begin < self.buf.len() && (self.buf[begin] <= 0x20) {
                    begin += 1;
                }
                if begin == self.buf.len() {
                    self.update_buf();
                    continue;
                }
                let mut res = 0;
                let (is_positive, mut end) = match self.buf[begin] {
                    b'+' => (true, begin + 1),
                    b'-' => (false, begin + 1),
                    _ => (true, begin),
                };
                while end < self.buf.len() && (self.buf[end] > 0x20) {
                    res = res * 10 + (self.buf[end] as i64 - '0' as i64);
                    end += 1;
                }
                if begin != self.buf.len() {
                    self.pos = end;
                    return if is_positive { res } else { -res };
                } else {
                    self.update_buf();
                }
            }
        }
    }
}

use input_mcr::*;
use std::cmp::*;

fn main() {
    input! {
        n: usize,
        m: usize,
    }
    let mut res = vec![];
    let left = (m + 1) / 2;
    let right = m - left;
    for i in 0..left {
        res.push((left - i, left + 1 + i));
    }
    for i in 0..right {
        res.push((left * 2 + right + 1 - i - 1, left * 2 + right + 1 + i + 1));
    }
    for &(a, b) in &res {
        println!("{} {}", a, b);
    }
}