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)

#include "bits/stdc++.h"
#include <unordered_set>
#define rep(i,n) for(int i = 0; i < n; i++)
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define vll vector<vector<long long>>
#define vl vector<long long>
#define vi vector<int>
#define vii vector<vector<int>>
#define pb push_back
#define pf push_front
#define ld long double
#define Sort(a) sort(a.begin(),a.end())
#define cSort(a,cmp) sort(a.begin(),a.end(),cmp)
#define reSort(a) sort(a.rbegin(), a.rend())
static const ll llMAX = numeric_limits<long long>::max();
static const int intMAX = numeric_limits<int>::max();
static const ll  llMIN = numeric_limits<long long>::min();
static const int intMIN = numeric_limits<int>::min();
static const ll d_5 = 100000;
static const ll d9_7 = 1000000007;
static const ll d_9 = 1000000000;
static const double PI=3.14159265358979323846;
template<class T>
T gcd(T a,T b){
  if(a==0){
    return b;

  }else if(b==0){
    return a;
  }
  while(1) {
    if(a < b) swap(a, b);
    if(!b) break;
    a %= b;
  }
  return a;
}
ll digitpower(ll a,ll b){//aのb乗を計算
  if(b==1){
    return a;
  }else if(b==0){
    return 1;
  }
  int mode=0;
  if(mode==0){
    if(b%2==1){
      ll tmp = digitpower(a,(b-1)/2);
      tmp%=d9_7;
      tmp*=tmp;
      tmp%=d9_7;
      tmp*=a;
      return (tmp)%d9_7;
    }else{
      ll tmp = digitpower(a,(b)/2);
      tmp%=d9_7;
      tmp*=tmp;
      tmp%=d9_7;
      return (tmp)%d9_7;
    }
  }else{
    if(b%2==1){
      ll tmp = digitpower(a,(b-1)/2);
      tmp*=tmp;
      tmp*=a;
      return (tmp);
    }else{
      ll tmp = digitpower(a,(b)/2);
      tmp*=tmp;
      return (tmp);
    }
  }
  return 0;
  
}
template<class T>
void Printvector(std::vector<T> &a){
  int size = a.size();
  rep(i,size){
    cout<<a[i]<<" ";
  }
  cout<<endl;
}
template<class T>
void Printvector(std::vector<std::vector<T> > &a){
  int size = a.size();
  rep(i,size){
    int size2=a[i].size();
    rep(j,size2){
      cout<<a[i][j]<<" ";
    }
    cout<<endl;
  }
  cout<<endl;
}
unordered_set<ll> primes(ll n){
  unordered_set<ll> ret;
  for(ll i=2;i*i<=n;i++){
    if(n%i==0){
      ret.insert(i);
      ret.insert(n/i);
    }
  }
  return ret;
}
long long modinv(long long a, long long m) {//modの逆元
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m; 
    if (u < 0) u += m;
    return u;
}
struct c{
  ll a;
  ll b;
};
template<class T>
vector<T> getaccum(vector<T> a){
  int size=a.size();
  vector<T> ans(size);
  ans[0]=a[0];
  for(int i=0;i<size-1;i++){
    ans[i+1]=ans[i]+a[i+1];
    //ans[i+1]%=d9_7;
  }
  return ans;
}
// 比較関数を定義
bool cmp( const c& left, const c& right ) {
    return left.b>right.b;
}

int main(void){
  ll n,m;
  cin>>n>>m;
  vii dd(m,vi(2));
  for(int i=0;i<m;i++){
    dd[i][0]=i;
    dd[i][1]=2*m-1-i;
  }
  int add=0;
  while(true){
    vi number(n);
    rep(i,n){
      number[i]=i;
    }
    int flag=0;
    set<pair<int,int>> matches;
    int cnt=0;
    rep(i,m){
      int n1=number[dd[i][0]+cnt];
      int n2=number[(dd[i][1]+cnt+add)%n];
      if(n1>n2){
        swap(n1,n2);
      }
      if(matches.find({n1,n2})!=matches.end()){
        flag=1;
        break;
      }else{
        matches.insert({n1,n2});
      }
    }
    if(flag==0){
      break;
    }else{
      add++;
    }
  }
  rep(i,m){
    cout<<dd[i][0]+1<<" "<<dd[i][1]+add+1<<endl;
  }
  return 0;
}

//clang++ -g -fsanitize=address -fno-omit-frame-pointer 1.cpp -o test.exe

//<<std::setprecision(30)

//重複削除
 /* std::sort(vec.begin(), vec.end());
 
  vec.erase(std::unique(vec.begin(), vec.end()), vec.end());*/

  //ペアの全探索
  /*do{
		int s=0;
		for(int i=0;i<n/2;i++)s+=v[i]^v[i+n/2];
		r=max(s,r);
		reverse(v.begin()+n/2,v.end());
	}while(next_permutation(v.begin(),v.end()));*/