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: Runtime Error)

use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: usize = itr.next().unwrap().parse().unwrap();

    // mod n
    let mut used = vec![false; n + 1000];
    let mut out = Vec::new();
    let mut i = 1;
    for idx in 1..n + 1 {
        if used[idx] {
            continue;
        }
        // mod n == iとなる組み合わせ
        let k1 = i;
        let k2 = n - i;
        used[i] = true;
        if idx + k1 <= n && !used[idx + k1] {
            writeln!(out, "{} {}", idx, idx + k1).ok();
            used[idx + k1] = true;
        } else if idx + k2 <= n && !used[idx + k2] {
            writeln!(out, "{} {}", idx, idx + k2).ok();
            used[idx + k2] = true;
        } else if idx > k2 && !used[i - k2] {
            writeln!(out, "{} {}", idx, idx - k2).ok();
            used[idx - k2] = true;
        } else if idx > k1 && !used[i - k1] {
            writeln!(out, "{} {}", idx, idx - k1).ok();
            used[idx - k1] = true;
        } else {
            continue;
        }
        i += 1;
        if i == m + 1 {
            break;
        }
    }
    stdout().write_all(&out).unwrap();
}