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)

//clear adj and visited vector declared globally after each test case
//check for long long overflow
//while adding and subs check if mod becomes -ve
//while using an integer directly in a builtin function add ll

#include <bits/stdc++.h>
#define int long long
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);cout.precision(dbl::max_digits10);
#define pb push_back
#define mod 1000000007 //998244353
#define lld long double
#define mii map<int, int>
#define mci map<char, int>
#define msi map<string, int>
#define pii pair<int, int>
#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()
#define rep(i,x,y) for(int i=x; i<y; i++)
#define fill(a,b) memset(a, b, sizeof(a))
#define vi vector<int>
#define setbits(x) __builtin_popcountll(x)
typedef std::numeric_limits< double > dbl;
using namespace std;
const long long N=2000005, INF=2000000000000000000;
int power(int a, int b, int p)
    {
        if(a==0)
        return 0;
        int res=1;
        a%=p;
        while(b>0)
        {
            if(b&1)
            res=(res*a)%p;
            b>>=1;
            a=(a*a)%p;
        }
        return res;
    }
int n,m,q,c=0;
vector <int> v[N];

void create(int x)
{
    int l=v[x].size();
    //cout<<x<<" yo "<<l<<"\n";
    /**if(l==n){cout<<x<<": ";
    for(int j=0;j<l;j++)
    cout<<v[x][j]<<" ";
    cout<<"\n";}**/
    if(l==n)
    return;
    if(l==0)
    {
        for(int i=1;i<=m;i++)
        {
            v[c].pb(i);
            create(c);
            c++;
        }
        return;
    }
    int b=v[x][l-1];
    v[x].pb(b);
    create(x);
    for(int i=b+1;i<=m;i++)
    {
        c++;
        for(int j=0;j<l;j++)
        v[c].pb(v[x][j]);
        v[c].pb(i);
        create(c);
    }
    return;
}
int32_t main()
{
    IOS;
    int n,m,z=0;
    cin>>n>>m;
    pii ar[m];
    for(int i=1;;i++)
    {
        ar[z]={i, n-i+1};
        z++;
        if(z==m)
        break;
    }
    for(int i=0;i<m;i++)
    cout<<ar[i].ff<<" "<<ar[i].ss<<"\n";
}