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>
#define ALL(a)  (a).begin(),(a).end()
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<long long, long long> Pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long> vll;
typedef vector<vector<long long>> vvll;
template <typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template <typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
const int INT_INF = 1 << 30;
#define MOD 1000000007LL
#define endl "\n"

int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll N, M;
  cin >> N >> M;
  int mmax = N / 2;
  if(N % 2 == 0) mmax--;
  int odd = mmax;
  int even = mmax;

  ll counter = 0;
  ll right = 0;
  //even
  for(int i = 0; i < even; i++){
    ll x = i+1, y = 2 * even - i;
    if(y > N) break;
    if(N % 2 == 0 && y - x == N / 2) continue;
    cout << x << ' ' << y << endl;
    right = y;
    counter++;
    if(counter == M) break;
  }

  if(counter == M) return 0;

  //odd
  for(int i = 0; i < odd; i++){
    ll x = right + i + 1, y = right + 2 * odd + 1 - i;
    if(y > N) break;
    if(N % 2 == 0 && y - x == N / 2) continue;
    cout << x << ' ' << y << endl;
    counter++;
    if(counter == M) break;
  }
  cout << odd << endl;
  cout << even << endl;
  cout << counter << endl;
}