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 <cassert>
#include <algorithm>
#include <array>
#include <bitset>
#include <cctype>
#include <cstdint>
#include <cmath>
#include <complex>
#include <chrono>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#include <random>
#include <memory>
#include <utility>
#include <limits>

/* template start */
 
#define rep(i, a, b) for (long long i = (a); (i) < (b); (i)++)
#define all(i) i.begin(), i.end()
#define debug(...) std::cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

void debug_out(){std::cerr<<std::endl;}

template<typename Head,typename... Tail>
void debug_out(Head h,Tail... t){
  std::cerr<<" "<<h<<":";
  debug_out(t...);
}
 
template <typename T1, typename T2>
std::ostream& operator<<(std::ostream& os, std::pair<T1, T2> pa) {
  return os << pa.first << " " << pa.second;
}
 
template <typename T>
std::ostream& operator<<(std::ostream& os, std::vector<T> vec) {
  for (std::size_t i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  return os;
}
 
template<typename T1,typename T2>
inline bool chmax(T1& a,T2 b){return a<b && (a=b,true);}
 
template<typename T1,typename T2>
inline bool chmin(T1& a,T2 b){return a>b && (a=b,true);}
 
long long pow_mod(long long a, long long b, long long mod=-1) {
  if(b==0)return 1;
  if ((a == 0)||(mod!=-1&&(a+mod)%mod==0))return 0;
  long long x = 1;
  while (b > 0) {
    if (b & 1)x = (mod!=-1)?(x * a) % mod:x*a;
    a = (mod!=-1)?(a * a) % mod:a*a;
    b >>= 1;
  }
  return x;
}

template <std::uint_fast64_t Modulus> class modint {
  using u64 = std::uint_fast64_t;

public:
  u64 a;

  constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
  constexpr u64 &value() noexcept { return a; }
  constexpr const u64 &value() const noexcept { return a; }
  constexpr modint operator+(const modint rhs) const noexcept {
    return modint(*this) += rhs;
  }
  constexpr modint operator-(const modint rhs) const noexcept {
    return modint(*this) -= rhs;
  }
  constexpr modint operator*(const modint rhs) const noexcept {
    return modint(*this) *= rhs;
  }
  constexpr modint operator/(const modint rhs) const noexcept {
    return modint(*this) /= rhs;
  }
  constexpr modint &operator+=(const modint rhs) noexcept {
    a += rhs.a;
    if (a >= Modulus) {
      a -= Modulus;
    }
    return *this;
  }
  constexpr modint &operator-=(const modint rhs) noexcept {
    if (a < rhs.a) {
      a += Modulus;
    }
    a -= rhs.a;
    return *this;
  }
  constexpr modint &operator*=(const modint rhs) noexcept {
    a = a * rhs.a % Modulus;
    return *this;
  }
  constexpr modint &operator/=(modint rhs) noexcept {
    u64 exp = Modulus - 2;
    while (exp) {
      if (exp % 2) {
        *this *= rhs;
      }
      rhs *= rhs;
      exp /= 2;
    }
    return *this;
  }
};

/* template end */

using ll = long long;
using P=std::pair<long long,long long>;

int main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);

  ll n,m;
  std::cin>>n>>m;

  rep(i,0,(m+1)/2){
    std::cout<<i+1<<" "<<m-i+1<<"\n";
  }

  rep(i,0,m/2){
    std::cout<<m+1+i+1<<" "<<m+1+(m-1-i)+1<<"\n";
  }

  return 0;
}