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: Time Limit Exceeded)

#include <bits/stdc++.h>
 
#define f(i,a,b) for( ll i = a; i < (ll) b ; i++ ) 
#define af(i,a,b) for( ll i = a; i >= (ll) b ; i--)
#define rep(i,a,b,k) for(ll i = a; i < b ; i+= k )
#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define sz(a) (ll) a.size()
#define all(a) a.begin(), a.end()
#define sor(a) sort( a.begin(), a.end() )
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define inter ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> ii ;
typedef vector<ll>  vi ;
typedef vector<ii> vii ;
 
const ll MAX= 2e6+100;


ll a[MAX];
int main(){
	fastio;
	ll n,m;
	cin>>n>>m;
	if(n&1){		
		f(i,1,m+1){
			cout<<i<<" "<<2*m+1-i<<endl;
		}
	}
	else{
		ll ind=0;
		f(i,1,m+1){
			while((a[ind%n] || a[(ind+i)%n])) ind++;
			a[ind]=a[(ind+i)%n]=1;
			cout<<ind%n+1<<" "<<(ind+i)%n+1<<endl;
		}
	}
	return 0;
}