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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ProblemE
{
    class Program
    {
        static void Main(string[] args)
        {
            var read = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            var N = read[0];
            var M = read[1];

            var half = M / 2;
            var endFirst = M%2 == 0 ? M + 1 : M;
            var startSecond = endFirst + 1;
            var endSecond = 2 * M + 1;
            for (int i = 0; i < half; i++)
            {
                Console.WriteLine((i + 1) + " " + (endFirst - i));
            }
            for (int i = 0; i < M - half; i++)
            {
                Console.WriteLine((startSecond + i) + " " + (endSecond - i));
            }
        }

        static int CalcDist(int a, int b, int N)
        {
            return Math.Min(b - a, a + N - b);
        }

        
//            #region odd case
//            if (N % 2 != 0)
//            {
//                for (int i = 0; i<M; i++)
//                {
//                    Console.WriteLine((i + 1) + " " + (N - i));
//                }
//}
//            #endregion 
//            else
//            {
//                for(int i=0;i<M;)
//                {
//                    var dists = new int[M];
//                    for (int j = 0; j<M; j++)
//                    {
//                        var dist = CalcDist(i + 1, N - i - j, N);
//                        if (!(N - j - 2 * i - 1 == N / 2) && !dists.Contains(dist))
//                        {
//                            Console.WriteLine((i + 1) + " " + (N - i - j));
//                            dists[i] = dist;
//                            i++;
//                            break;
//                        }
//                    }
//                }
//            }
    }
}