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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;

namespace AtCoder
{
    static class Program_E
    {
        class Solver
        {

            public void Solve()
            {
                var nm = Ints();
                var n = nm[0];
                var m = nm[1];
                for(int i =1;i<=m;++i)
                {
                    Console.WriteLine(string.Format("{0} {1}",i,2*m-i+1));
                }


            }



            // ----------------------------------------------------------------------------

            public int Int()
            {
                return int.Parse(Console.ReadLine());
            }
            public int[] Ints()
            {
                return Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
            }
            public long[] Longs()
            {
                return Console.ReadLine().Split(' ').Select(s => long.Parse(s)).ToArray();
            }
            public int[][] MultiInts(int num)
            {
                var result = new int[num][];
                for (int i = 0; i < num; ++i)
                {
                    result[i] = Ints();
                }
                return result;
            }
            public T[] MultiItems<T>(int num, Func<string, T> parser)
            {
                var result = new T[num];
                for (int i = 0; i < num; ++i)
                {
                    result[i] = parser(Console.ReadLine());
                }
                return result;
            }
        }



        static void Main(string[] args)
        {
            new Solver().Solve();
        }
    }
    struct ModInt
    {
        const int MOD = 1000000007;
        long Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public ModInt(long data) { if ((Data = data % MOD) < 0) Data += MOD; }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator long(ModInt modInt) => modInt.Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator ModInt(long val) => new ModInt(val);
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, ModInt b)
        {
            long res = a.Data + b.Data;
            return new ModInt() { Data = res >= MOD ? res - MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, long b) => a.Data + b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, ModInt b)
        {
            long res = a.Data - b.Data;
            return new ModInt() { Data = res < 0 ? res + MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, long b) => a.Data - b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator *(ModInt a, ModInt b) => new ModInt() { Data = a.Data * b.Data % MOD };
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator /(ModInt a, ModInt b) => a.Data * GetInverse(b);

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public override string ToString() => Data.ToString();

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        static long GetInverse(long a)
        {
            long div;
            long p = MOD;
            long x1 = 1, y1 = 0, x2 = 0, y2 = 1;
            while (true)
            {
                if (p == 1) return x2;
                div = a / p;
                x1 -= x2 * div;
                y1 -= y2 * div;
                a %= p;
                if (a == 1) return x1;
                div = p / a;
                x2 -= x1 * div;
                y2 -= y1 * div;
                p %= a;
            }
        }
    }

}