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)

use std::cmp;
use std::collections::*;

pub struct Scanner {
    line: Vec<String>,
    idx: usize,
}

#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner {
        return Scanner {
            line: vec![],
            idx: 0,
        };
    }
    fn next<T: std::str::FromStr>(&mut self) -> T {
        if self.idx == self.line.len() {
            let mut buffer = String::new();
            std::io::stdin().read_line(&mut buffer).ok();
            let a: Vec<&str> = buffer.trim().split(' ').collect();
            self.line = a.iter().map(|&s| String::from(s)).collect();
            self.idx = 0;
        }
        let ret: T = self.line[self.idx].parse::<T>().ok().expect("Parse error.");
        self.idx += 1;
        ret
    }
    fn next_string(&mut self) -> Vec<char> {
        let s: String = self.next();
        return s.chars().collect::<Vec<char>>();
    }
    fn next_vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        let mut v: Vec<T> = Vec::with_capacity(n);
        for _i in 0..n {
            v.push(self.next());
        }
        v
    }
}



fn main(){
    let mut scan = Scanner::new();
    let n: i32 = scan.next();
    let m: usize = scan.next();
    

    if n % 2 == 1{
       for i in 0..m {
            println!("{} {}", i + 1 , n - i as i32);
        }
    } else{
        let mut v: Vec<(i32, i32)> = Vec::new();

        let mut i = 0;
        let mut len = n/2-1;
        while len > 0{
            v.push((i, i+len));
            i += 1;
            len -= 2;
        }
        i = n/2;
        len = n/2 - 2;
        while len > 0{
            v.push((i, i+len));
            i += 1;
            len -= 2;
        }
        for i in 0..m {
            println!("{} {}", v[i].0 + 1 , v[i].1 + 1);
        }
    }
}