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)

#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("unroll-loops,rename-registers")
//#pragma GCC target("sse4,avx,avx2")

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <iomanip>
#include <chrono>
#include <random>
#include <string>
#include <queue>
#include <tuple>
#include <unordered_map>
#include <sstream>
#include <numeric>
#include <complex>
#include <immintrin.h>
#include <unordered_set>

using namespace std;

using ll = long long;
using ld = long double;
using VectorType = ld;
using ull = unsigned long long;

const ll INF = 1e15;
const ld eps = 1e-7;
const ld pi = atan2(0, -1);
const ll mod = 1e9 + 7;
const ll P = 179, Q = 239;
const ll A = 26;

mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());

ll sign(ll a) {
	if (a < 0) return -1;
	if (a == 0) return 0;
	return 1;
}

bool equal(ld a, ld b) {
	return abs(a - b) < eps;
}

struct query {
	ll l, r, i;

	query(ll l = 0, ll r = 0, ll i = 0) : l(l), r(r), i(i) {};
};

bool mini(ll &a, ll b) {
	if (b < a) {
		a = b;
		return true;
	}
	return false;
}

bool maxi(ll &a, ll b) {
	if (b > a) {
		a = b;
		return true;
	}
	return false;
}

bool mini(ld &a, ld b) {
	if (b < a) {
		a = b;
		return true;
	}
	return false;
}

ll mmax(ll a, ll b) {
	if (a > b) return a;
	return b;
}

ll mmin(ll a, ll b) {
	if (a < b) return a;
	return b;
}

ld mmax(ld a, ld b) {
	if (a > b) return a;
	return b;
}

ld mmin(ld a, ld b) {
	if (a < b) return a;
	return b;
}

bool maxi(ld &a, ld b) {
	if (b > a) {
		a = b;
		return true;
	}
	return false;
}

ll sub(ll a, ll b) {
	a -= b;
	if (a < 0) {
		a += mod;
	}
	return a;
}

ll add(ll a, ll b) {
	a += b;
	if (a > mod) {
		a -= mod;
	}
	return a;
}

ll mul(ll a, ll b) {
	return (a * b) % mod;
}

ll n, m;
ll k;
vector <vector <ll>> vec;
vector <bool> used;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cin >> n >> m;
	for (int i = 0; i < m; ++i) {
		cout << i + 1 << ' ' << n - i << '\n';
	}
	return 0;
}