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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

const mod = 998244353
const facNum = 300001

func main() {
	sc := newScanner(os.Stdin)
	N, M := sc.i(), sc.i()
	harf := N / 2
	battleRooms := make([][]int, 0, M)
	for i := 0; i < M; i++ {
		first := harf - i
		second := harf + 1 + i
		battleRooms = append(battleRooms, []int{first, second})
	}
	for _, room := range battleRooms {
		fmt.Printf("%d %d\n", room[0], room[1])
	}
}

// fmt.Println("Yes")
// fmt.Println("No")

// I/O
type scanner struct {
	sc *bufio.Scanner
}

func newScanner(input io.Reader) *scanner {
	sc := bufio.NewScanner(input)
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 1024), int(1e+9))
	return &scanner{sc}
}

func (s *scanner) s() string {
	s.sc.Scan()
	return s.sc.Text()
}

func (s *scanner) i() int {
	i, e := strconv.Atoi(s.s())
	if e != nil {
		panic(e)
	}
	return i
}

func (s *scanner) f() float64 {
	f, e := strconv.ParseFloat(s.s(), 64)
	if e != nil {
		panic(e)
	}
	return f
}

func (s *scanner) bs() []byte {
	return []byte(s.s())
}

func (s *scanner) is(n int) []int {
	res := make([]int, n)
	for i := 0; i < n; i++ {
		res[i] = s.i()
	}
	return res
}

func (s *scanner) fs(n int) []float64 {
	res := make([]float64, n)
	for i := 0; i < n; i++ {
		res[i] = s.f()
	}
	return res
}

//const factorial = new
func Pow(a, n int) int {
	ans := 1
	for n > 0 {
		if (n & 1) == 1 {
			ans = ans * a % mod
		}
		a = a * a % mod
		n = n >> 1
	}
	return ans
}

func Gcd(a, b int) int {
	if a < b {
		return Gcd(b, a)
	}
	for b != 0 {
		a, b = b, a%b
	}
	return a
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func combination(n, k int, fac, ifac []int) int {
	if n < k || n < 0 {
		return 0
	}
	if k == 0 {
		return 1
	}
	//while n != 0
	ans := ifac[k] * ifac[n-k] % mod
	return ans * fac[n] % mod
}

// 階乗: factorial
// コンビネーションを計算する際に前もって計算しておく
// fac[k] => k! (mod M)
// ifac[k] => k!^(M-2) (mod M)
// n: ex. 2 * 10^5 => 200001
// fac, ifac := factorial()
func factorial() (fac []int, ifac []int) {
	fac = make([]int, facNum)
	fac[0] = 1
	ifac = make([]int, facNum)
	ifac[0] = 1
	for i := 1; i < facNum; i++ {
		fac[i] = fac[i-1] * i % mod
		ifac[i] = ifac[i-1] * Pow(i, mod-2) % mod
	}
	return
}