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: Accepted)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<string> vs;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;
struct Edge {
	ll to;
 	ll cost;
 };
using Graph = vector<vector<Edge>>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
/***** define constant start *******/
const ll MOD = 1000000007;
const double PI = 3.1415926535897932;
const ll INF = LLONG_MAX;
string abc = "abcdefghijklmnopqrstuvwxyz";
string ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll MAX = 10000000;
/****** define constant end ********/
void recursive_comb(int *indexes, int s, int rest, std::function<void(int *)> f) {
  if (rest == 0) {
    f(indexes);
  } else {
    if (s < 0) return;
    recursive_comb(indexes, s - 1, rest, f);
    indexes[rest - 1] = s;
    recursive_comb(indexes, s - 1, rest - 1, f);
  }
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(int n, int k, std::function<void(int *)> f) {
  int indexes[k];
  recursive_comb(indexes, n - 1, k, f);
}
/****** define variable start ******/
ll n, m;
/******* define variable end *******/
int solve() {
	cin >> n >> m;
	vector<pair<ll, ll>> ans;
	if(m % 2 == 0){
		ll waruni = (m * 2 + 1) / 2 + 1;
		for(ll i = 1;;i++){
			if(waruni-i+1 - i <= 1){
				break;
			}
			ans.push_back(make_pair(i, waruni-i+1));
		}
		for(ll i = waruni+ 1;;i++){
			if(m * 2 + 1 - (i - waruni) + 1 - i <= 0){
				break;
			}
			ans.push_back(make_pair(i, m * 2 + 1 - (i - waruni) + 1));
		}
	}else{
		ll waruni = (m * 2 + 1) / 2;
		for(ll i = 1;;i++){
			if(waruni-i+1 - i <= 1){
				break;
			}
			ans.push_back(make_pair(i, waruni-i+1));
		}
		for(ll i = waruni+ 1;;i++){
			if(m * 2 + 1 - (i - waruni) + 1 - i <= 0){
				break;
			}
			ans.push_back(make_pair(i, m * 2 + 1 - (i - waruni) + 1));
		}
	}
	REP(i, m){
		cout << ans[i].first << " " << ans[i].second << endl;
	}
	return 0;
}

int main() {
	//clock_t start = clock();
	cout << std::fixed << std::setprecision(10);
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int end_solve = solve();
	/*
	clock_t end = clock();
	const double time = static_cast<double>(end - start) / CLOCKS_PER_SEC *
	1000.0; printf("time %lf[ms]\n", time);
	*/

	return end_solve;
}