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 C# (Mono-csc 3.5.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Console;

namespace AtCoder_1
{
    class Program
    {
        static void Main(string[] args)
        {
            var NM = CinTAr();
            int its = 1;
            for(int i = 1; i <= NM[1]; i++)
            {
                int b = NM[0] + 1 - its;
                Cout($"{its} {b}");
                its++;
            }


            Cin();
        }

        static Random Rand = new Random();
        static string[] AlphabetTable = new string[] { "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z" };
        static int GetDigits(long num)
        {
            return (int)Math.Floor(Math.Log10(num)) + 1;
        }
        static int GetNumByDigit(long num, int digit)
        {
            return (int)((num / ((int)Math.Pow(10, digit - 1))) % 10);
        }
        static int loop(int num, int max)
        {
            if (num > max)
            {
                return num % max;
            }
            else if (num < 0)
            {
                return max - (-num) % max;
            }
            else
            {
                return num;
            }
        }
        static long GCD(long a, long b)
        {
            if (a < b) { swap(ref a, ref b); }
            if (a % b == 0) { return b; }
            else { return GCD(b, a % b); }
        }

        static long LCM(long a, long b)
        {
            return a * b / GCD(a, b);
        }

        static void swap<T>(ref T a, ref T b)
        {
            T t = a;
            a = b;
            b = t;
        }

        static readonly int MOD10_9_7 = (1000000000 + 7);
        static bool IsInt(double d) { return (int)d == d; }
        static string Cin() { return ReadLine(); }
        static string[] CinAr() { return ReadLine().Split(' '); }
        static int CinT() { return S2i(Cin()); }
        static int[] CinTAr() { return CinAr().Select((x, i) => { return S2i(x); }).ToArray(); }
        static long CinL() { return S2L(Cin()); }
        static long[] CinLAr() { return CinAr().Select((x, i) => { return S2L(x); }).ToArray(); }
        static double CinD() { return S2D(Cin()); }
        static double[] CinDAr() { return CinAr().Select((x, i) => { return S2D(x); }).ToArray(); }
        static int S2i(string s) { return int.Parse(s); }
        static long S2L(string s) { return long.Parse(s); }
        static double S2D(string s) { return double.Parse(s); }
        static void Cout(object s) { WriteLine(s); }
        static long[] yakusuu(long n)
        {
            if (n == 1) { return new long[] { 1 }; }
            List<long> res = new List<long>();
            res.Add(1);
            for (long i = 2; i <= n / 2 + 1; i++)
            {
                if (n % i == 0)
                {
                    res.Add(i);
                }
            }
            res.Add(n);
            return res.ToArray();
        }
        static bool isPrime(long num)
        {
            if (num == 1) { return false; }
            for (long i = 2; i <= Math.Sqrt(num) + 1; i++)
            {
                if (num % i == 0) { return false; }
            }
            return true;
        }

        static void CoutAr<Type>(IEnumerable<Type> types)
        {
            WriteLine(string.Join(" ", types));
        }

        public static long nCr(int n, int r)
        {
            // naive: return Factorial(n) / (Factorial(r) * Factorial(n - r));
            return nPr(n, r) / Factorial(r);
        }

        public static long nPr(int n, int r)
        {
            // naive: return Factorial(n) / Factorial(n - r);
            return FactorialDivision(n, n - r);
        }

        private static long FactorialDivision(int topFactorial, int divisorFactorial)
        {
            long result = 1;
            for (int i = topFactorial; i > divisorFactorial; i--)
                result *= i;
            return result;
        }

        private static long Factorial(int i)
        {
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);
        }
    }

    public struct Pair<T>
    {
        public T a;
        public T b;
        public T c;
        public Pair(T _a, T _b) { a = _a; b = _b; c = default(T); }
        public Pair(T _a, T _b, T _c) { a = _a; b = _b; c = _c; }
    }
}