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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

/*****Author: Satyajeet Singh, Delhi Technological University************************************/
    import java.io.*;
    import java.util.*;
    import java.text.*; 
    import java.lang.*;
    import java.math.*;
public class Main{
/*********************************************Constants******************************************/
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));        
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static long mod=(long)1e9+7;
    static long mod1=998244353;
    static ArrayList<Integer> graph[];
    static int pptr=0,pptrmax=0;
    static String st[];
/*****************************************Solution Begins***************************************/
    public static void main(String args[]) throws Exception{
        int n=pi();
        int m=pi();
        for(int i=2;i<m+2;i++){
            out.println(1+" "+i);
        }
/****************************************Solution Ends**************************************************/
        out.flush();
        out.close();
    }
/****************************************Template Begins************************************************/
    static void nl() throws Exception{
        pptr=0;
        st=br.readLine().split(" ");
        pptrmax=st.length;
    }
    static void nls() throws Exception{
        pptr=0;
        st=br.readLine().split("");
        pptrmax=st.length;
    }
    static int pi() throws Exception{
        if(pptr==pptrmax)
            nl();
        return Integer.parseInt(st[pptr++]);
    }
    static long pl() throws Exception{
        if(pptr==pptrmax)
            nl();
        return Long.parseLong(st[pptr++]);
    }
    static double pd() throws Exception{
        if(pptr==pptrmax)
            nl();
        return Double.parseDouble(st[pptr++]);
    }
    static String ps() throws Exception{
        if(pptr==pptrmax)
            nl();
        return st[pptr++];
    }
/***************************************Precision Printing**********************************************/
    static void printPrecision(double d){
        DecimalFormat ft = new DecimalFormat("0.000000"); 
        out.print(ft.format(d));
    }
/**************************************Bit Manipulation**************************************************/
    static int countBit(long mask){
        int ans=0;
        while(mask!=0){
            mask&=(mask-1);
            ans++;
        }
        return ans;
    }
/******************************************Graph*********************************************************/
    static void Makegraph(int n){
        graph=new ArrayList[n];
        for(int i=0;i<n;i++)
            graph[i]=new ArrayList<>();
    }
    static void addEdge(int a,int b){
        graph[a].add(b);
    }
    // static void addEdge(int a,int b,int c){
    //     graph[a].add(new Pair(b,c));
    // }    
/*********************************************PAIR********************************************************/
    static class Pair{
        int u;
        int v;
        public Pair(int u, int v) {
            this.u = u;
            this.v = v;
        }
        public int hashCode() {
            int hu = (int) (u ^ (u >>> 32));
            int hv = (int) (v ^ (v >>> 32));
            return 31 * hu + hv;
        }
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return u == other.u && v == other.v;
        }
        public String toString() {
            return "[u=" + u + ", v=" + v + "]";
        }
    }
/******************************************Long Pair*******************************************************/
    static class Pairl{
        long u;
        long v;
        public Pairl(long u, long v) {
            this.u = u;
            this.v = v;
        }
        public int hashCode() {
            int hu = (int) (u ^ (u >>> 32));
            int hv = (int) (v ^ (v >>> 32));
            return 31 * hu + hv;
        }
        public boolean equals(Object o) {
            Pairl other = (Pairl) o;
            return u == other.u && v == other.v;
        }
        public String toString() {
            return "[u=" + u + ", v=" + v + "]";
        }
    }
/*****************************************DEBUG***********************************************************/
    public static void debug(Object... o){
        System.out.println(Arrays.deepToString(o));
    }
/************************************MODULAR EXPONENTIATION***********************************************/
    static long modulo(long a,long b,long c){
        long x=1,y=a%c;
        while(b > 0){
            if(b%2 == 1)
                x=(x*y)%c;
            y = (y*y)%c;
            b = b>>1;
        }
        return  x%c;
    }
/********************************************GCD**********************************************************/
    static long gcd(long x, long y){
        if(x==0)
            return y;
        if(y==0)
            return x;
        long r=0, a, b;
        a = (x > y) ? x : y; 
        b = (x < y) ? x : y;
        r = b;
        while(a % b != 0){
            r = a % b;
            a = b;
            b = r;
        }
        return r;
    }
}