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;

#define rep(i,s,n) for (int i = (int)s; i < (int)n; i++)
#define ll long long
#define ld long double
#define pb push_back
#define eb emplace_back
#define All(x) x.begin(), x.end()
#define Range(x, i, j) x.begin() + i, x.begin() + j
#define lbidx(x, y) lower_bound(x.begin(), x.end(), y) - x.begin()
#define ubidx(x, y) upper_bound(x.begin(), x.end(), y) - x.begin()
#define llbidx(x, y, z) lower_bound(x.begin(), x.end(), z) - lower_bound(x.begin(), x.end(), y) // 二要素間の距離
#define deg2rad(deg) ((((double)deg)/((double)360)*2*M_PI))
#define rad2deg(rad) ((((double)rad)/(double)2/M_PI)*(double)360)
#define Find(set, element) set.find(element) != set.end()
#define Decimal(x) printf("%.10f\n", x) // 小数点を10桁まで表示
// debug用
#define PrintVec(x) for (auto elementPrintVec: x) { cout << elementPrintVec << " "; } cout << "\n";
#define debug(x) cerr << #x << ": " << (x) << "\n";
#define endl "\n"
// gcj print用
#define Case(x) printf("Case #%d: ", x);

typedef pair<int, int> PI;
typedef pair<ll, ll> PLL;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef vector<vector<vector<int>>> vvvl;
typedef vector<PI> vpi;
typedef vector<vector<PI>> vvpi;
typedef vector<vector<vector<PI>>> vvvpi;
typedef vector<PLL> vpl;
typedef vector<vector<PLL>> vvpl;
typedef vector<vector<vector<PLL>>> vvvpl;

int POWINT(int x, int n) {
  int ret = 1;
  while (n > 0) {
    if (n & 1) ret *= x;
    x *= x;
    n >>= 1;
  }
  return ret;
};

ll POWLL(ll x, int n) {
  ll ret = 1;
  while (n > 0) {
    if (n & 1) ret *= x;
    x *= x;
    n >>= 1;
  }
  return ret;
};

template<class T>
inline bool chmax(T &a, T b) {
  if(a < b) {
    a = b;
    return true;
  }
  return false;
};

template<class T>
inline bool chmin(T &a, T b) {
  if(a > b) {
    a = b;
    return true;
  }
  return false;
};

void print() {
  cout << endl;
}

template <class T>
void print(vector<T> &vec) {
  for (auto& a : vec) {
    cout << a;
    if (&a != &vec.back()) cout << " ";
  }
  cout << endl;
}

template <class T>
void print(vector<T> &vec, ll k){
   ll n = vec.size();
   k = min(k, n);
   rep(i, 0, k-1) cout << vec[i] << " ";
   cout << vec[k-1] << endl;
}

template <class T>
void print(vector<vector<T>> &df) {
  for (auto& vec : df) {
    print(vec);
  }
}

template<class T, class U>
void print(pair<T,U> &p){
  cout << p.first << " " << p.second << "\n";
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N, M;
  cin >> N >> M;

  if (M & 1) {
    int m = M;
    int half_M = (M+1)/2;
    for (int l = 1; l <= half_M; l++) {
      if (m <= 0) break;
      int r = M+2-l;
      cout << l << " " << r << endl;
      m--;
    }

    if (m <= 0) return 0;

    for (int l = M+2; ;l++) {
      if (m <= 0) break;
      int r = (2*M+1) + (M+2) - l;
      cout << l << " " << r << endl;
      m--;
    }
  } else {
    int m = M;
    int half_M = M / 2;
    rep(l, 1, half_M+1) {
      if (m <= 0) break;
      int r = M+1-l;
      cout << l << " " << r << endl;
      m--;
    }

    if (m <= 0) return 0;

    for (int l = M+1; ;l++) {
      if (m <= 0) break;
      int r = 2 * M + 1 + M + 1 - l;
      cout << l << " " << r << endl;
      m--;
    }
  }


  return 0;
};