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)

#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
#include <random>
using namespace std;
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
typedef vector<int> vi;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod = 1000000007;
int rnd(int x) { return mrand() % x; }
ll powmod(ll a, ll b)
{
	ll res = 1;
	a %= mod;
	assert(b >= 0);
	for (; b; b >>= 1)
	{
		if (b & 1)
			res = res * a % mod;
		a = a * a % mod;
	}
	return res;
}
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

static uint MD = mod;
struct ModInt
{
	using M = ModInt;
	uint v;
	ModInt(ll _v = 0) { set_v(uint(_v % MD + MD)); }
	M &set_v(uint _v)
	{
		v = (_v < MD) ? _v : _v - MD;
		return *this;
	}
	explicit operator bool() const { return v != 0; }
	M operator-() const { return M() - *this; }
	M operator+(const M &r) const { return M().set_v(v + r.v); }
	M operator-(const M &r) const { return M().set_v(v + MD - r.v); }
	M operator*(const M &r) const { return M().set_v(uint(ull(v) * r.v % MD)); }
	M operator/(const M &r) const { return *this * r.inv(); }
	M &operator+=(const M &r) { return *this = *this + r; }
	M &operator-=(const M &r) { return *this = *this - r; }
	M &operator*=(const M &r) { return *this = *this * r; }
	M &operator/=(const M &r) { return *this = *this / r; }
	bool operator==(const M &r) const { return v == r.v; }
	M pow(ll n) const
	{
		M x = *this, r = 1;
		while (n)
		{
			if (n & 1)
				r *= x;
			x *= x;
			n >>= 1;
		}
		return r;
	}
	M inv() const { return pow(MD - 2); }
	friend ostream &operator<<(ostream &os, const M &r) { return os << r.v; }
};
using Mint = ModInt;
//head
// cin >> t;
// for (int cas = 1; cas <= t; ++cas)
// {
// 	cout << "Case #" << cas << ": ";
// }
int main()
{
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	int l=1,r=n;
	rep(i,0,m) {
		cout<<l<<" "<<r<<endl;
		l++; r--;
	}
	return 0;
}