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;
#if __has_include(<boost/multiprecision/cpp_int.hpp>)
	#include <boost/multiprecision/cpp_int.hpp>
	#include <boost/multiprecision/cpp_dec_float.hpp>
	using bll = boost::multiprecision::cpp_int;
	using bdouble = boost::multiprecision::cpp_dec_float_100;
	using namespace boost::multiprecision;
#endif
#ifdef LOCAL_DEV
	void debug_impl() { std::cerr << '\n'; }
	template<typename Head, typename... Tail> void debug_impl(Head head, Tail... tail) { std::cerr << " " << head << (sizeof...(tail) ? "," : ""); debug_impl(tail...); }
	#define debug(...) do { std::cerr << "(" << #__VA_ARGS__ << ") ="; debug_impl(__VA_ARGS__); } while (false)
#else
	#define debug(...) do {} while (false)
#endif
#ifdef LOCAL_TEST
	#define BOOST_STACKTRACE_USE_ADDR2LINE
	#define BOOST_STACKTRACE_ADDR2LINE_LOCATION /usr/local/opt/binutils/bin/addr2line
	#define _GNU_SOURCE
	#include <boost/stacktrace.hpp>
	namespace std {
		template<typename T> class dvector : public std::vector<T> {
		public:
			dvector() : std::vector<T>() {}
			explicit dvector(size_t n, const T& value = T()) : std::vector<T>(n, value) {}
			dvector(const std::vector<T>& v) : std::vector<T>(v) {}
			dvector(const std::initializer_list<T> il) : std::vector<T>(il) {}
			dvector(const std::string::iterator first, const std::string::iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::iterator first, const typename std::vector<T>::iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::reverse_iterator first, const typename std::vector<T>::reverse_iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::const_iterator first, const typename std::vector<T>::const_iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::const_reverse_iterator first, const typename std::vector<T>::const_reverse_iterator last) : std::vector<T>(first, last) {}
			T& operator[](size_t n) {
				try { return this->at(n); } catch (const std::exception& e) {
					std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n);
				}
			}
			const T& operator[](size_t n) const {
				try { return this->at(n); } catch (const std::exception& e) {
					std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n);
				}
			}
		};
	}
	class dbool {
	private:
		bool boolvalue;
	public:
		dbool() : boolvalue(false) {}
		dbool(bool b) : boolvalue(b) {}		
		dbool(const dbool&) = default;
		dbool(dbool&&) = default;
		dbool& operator=(const dbool&) = default;
		dbool& operator=(dbool&&) = default;
		operator bool&() { return boolvalue; }
		operator const bool&() const { return boolvalue; }
	};
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::dvector<T>& v) {
		for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::dvector<std::dvector<T>>& vv) {
		s << "\\\n"; for (size_t i = 0; i < vv.size(); ++i){ s << vv[i] << "\n"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::deque<T>& v) {
		for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::set<T>& se) {
		s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::multiset<T>& se) {
		s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
	template <typename T, size_t N> std::ostream& operator<<(std::ostream& s, const std::array<T, N>& a) {
		s << "{ "; for (size_t i = 0; i < N; ++i){ s << a[i] << "\t"; } s << "}"; return s; }
	template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::map<T1, T2>& m) {
		s << "{\n"; for (auto itr = m.begin(); itr != m.end(); ++itr){ s << "\t" << (*itr).first << " : " << (*itr).second << "\n"; } s << "}"; return s; }
	template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::pair<T1, T2>& p) {
		return s << "(" << p.first << ", " << p.second << ")"; }
	#define vector dvector
	#define bool dbool
	class SIGFPE_exception : std::exception {};
	class SIGSEGV_exception : std::exception {};
	void catch_SIGFPE([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGFPE_exception(); }
	void catch_SIGSEGV([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGSEGV_exception(); }
	signed convertedmain();
	signed main() { signal(SIGFPE, catch_SIGFPE); signal(SIGSEGV, catch_SIGSEGV); return convertedmain(); }
	#define main() convertedmain()
#endif
//#define int long long
using ll = long long;
//constexpr int INF = 1e9;//INT_MAX=(1<<31)-1=2147483647
constexpr ll INF = (ll)1e18;//(1LL<<63)-1=9223372036854775807
constexpr ll MOD = (ll)1e9 + 7;
constexpr double EPS = 1e-9;
constexpr ll dx[4] = {1LL, 0LL, -1LL, 0LL};
constexpr ll dy[4] = {0LL, 1LL, 0LL, -1LL};
constexpr ll dx8[8] = {1LL, 0LL, -1LL, 0LL, 1LL, 1LL, -1LL, -1LL};
constexpr ll dy8[8] = {0LL, 1LL, 0LL, -1LL, 1LL, -1LL, 1LL, -1LL};
#define rep(i, n)   for(ll i=0, i##_length=(n); i< i##_length; ++i)
#define repeq(i, n) for(ll i=1, i##_length=(n); i<=i##_length; ++i)
#define rrep(i, n)   for(ll i=(n)-1; i>=0; --i)
#define rrepeq(i, n) for(ll i=(n)  ; i>=1; --i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
void p() { std::cout << '\n'; }
template<typename Head, typename... Tail> void p(Head head, Tail... tail) { std::cout << head << (sizeof...(tail) ? " " : ""); p(tail...); }
template<typename T> inline void pv(std::vector<T>& v) { for(ll i=0, N=v.size(); i<N; i++) std::cout << v[i] << " \n"[i==N-1]; }
template<typename T> inline bool chmax(T& a, T b) { return a < b && (a = b, true); }
template<typename T> inline bool chmin(T& a, T b) { return a > b && (a = b, true); }
template<typename T> inline void uniq(std::vector<T>& v) { v.erase(std::unique(v.begin(), v.end()), v.end()); }

/*-----8<-----template-----8<-----*/

/*-----8<-----library-----8<-----*/

constexpr ll N_MAX = 100000;

ll N, M;

void solve() {
	scanf("%lld%lld", &N, &M);
	
	
	vector<pair<ll,ll>> ans;
	if(N%2==0){
		ll dis=N-1;
		ll start=1;
		repeq(i,(M+1)/2){
			ans.push_back({start, start+dis});

			dis-=2;
			start++;
		}
		dis--;
		//repeq(i,M/2){
		while(ans.size()<M){
			ll j=start;
			ans.push_back({j, j+dis});
			dis-=2;
			start++;
		}		
	}else{
		ll dis=N-1;
		repeq(i,M){
			ans.push_back({i, i+dis});
			dis-=2;
		}
	}



	
	rep(i,M)printf("%lld %lld\n", ans[i].first,ans[i].second);
	
}

// https://atcoder.jp/contests/abc165/tasks/abc165_e
signed main() {
	solve();
	return 0;
}