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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int  long long int
#define mod 1000000000000007
#define pb push_back
#define mp make_pair
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define gcd __gcd
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define all(i) i.begin(),i.end()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define set_bit __builtin_popcount //gives number of set_bits
#define ff first
#define ss second
#define cases(t) int t;cin>>t; rep(casn,1,t+1)
#define el cout<<"\n";
#define deci(n) int n; scanf("%lld",&n);
#define decs(n) string n; cin>>n;
#define sz(v) v.size()
#define repit(m) for(auto it=m.begin();it!=m.end();it++)
// #define find(m,i) (m.find(i)!=m.end())
// #define find(m,i) (m.find(i)==m.end())
#define cYES cout<<"YES\n";
#define cNO cout<<"NO\n";
#define ins insert
#define imax 1000000000000000000
#define imin -1000000000000000000
// #define NP 50000
//  cerr << "\nTime : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n";
 
//  IIITA
//SUNEET KUMAR -->  ਸੁਨੀਤ
 
// scanning numbers extremely fast/////////
void fs(int &number)
{
    bool negative = false;
    register int c;
    number = 0;
    c = getchar();
    if (c=='-')
    {
        negative = true;
        c = getchar();
    }
    for (; (c>47 && c<58); c=getchar())
        number = number *10 + c - 48;
    if (negative)
        number *= -1;
}
// ========================================
int power(int a,int x)
{
        a%=mod;
    int res=1;
    while(x>0)
    {
        if(x&1)
        res=res*a%mod;
        a=a*a%mod;
        x>>=1;
    }
    return res;
}
int max(int a,int b)
{
    if(a>b)
    {
        return a;
    }
    else
    {
        return b;
    }
}
int min(int a,int b)
{
    if(a>b)
    {
        return b;
    }
    else
    {
        return a;
    }
}
int madd(int a,int b)
{
    int ans;
    ans=(a%mod+b%mod)%mod;
    return ans;
}
int mmult(int a,int b)
{
    int ans=1;
    ans=((a%mod)*(b%mod))%mod;
    return ans%mod;
}
int modinverse(int a)
{
    return power(a,mod-2);
}
 
void nhi_degi()
{
    int n,m;cin>>n>>m;
    int count=1;
    // vector<pair<int,int>>v;
    for(int i=1;i<=m;i++)
    {
        cout<<1<<" "<<1+i;el
    }
}
signed main()
{
    // cases(t)
    // {
        nhi_degi();
    // }
}