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>
#include <chrono> 
 
using namespace std;
using namespace std::chrono; 
 
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #pragma optimization_level 3
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define f0r(a, b) for (long long a = 0; a < (b); ++a)
#define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
#define f0rd(a, b) for (long long a = (b); a >= 0; --a)
#define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define pb push_back
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define fix(prec) {cout << setprecision(prec) << fixed;}
#define mp make_pair
#define f first
#define s second
#define presum(p, a, n) {p[0] = a[0]; for (int i = 1; i < (n); i++) p[i] = a[i] + p[i-1];}
#define all(v) v.begin(), v.end()
#define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
#define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
#define ai(a, n) for (int ele = 0; ele < n; ele++) cin >> a[ele];
#define ain(a, lb, rb) for (int ele = lb; ele <= rb; ele++) cin >> a[ele];
#define ao(a, n) {for (int ele = 0; ele < (n); ele++) { if (ele) cout << " "; cout << a[ele]; } cout << '\n';}
#define aout(a, lb, rb) {for (int ele = (lb); ele <= (rb); ele++) { if (ele > (lb)) cout << " "; cout << a[ele]; } cout << '\n';}
typedef long long ll;
typedef double ld;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
 
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
  cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
 
mt19937 rng(steady_clock::now().time_since_epoch().count());
// mt19937 rng(61378913);
/* usage - just do rng() */
 
void usaco(string filename) {
  // #pragma message("be careful, freopen may be wrong")
	freopen((filename + ".in").c_str(), "r", stdin);
	freopen((filename + ".out").c_str(), "w", stdout);
}
 
const lld pi = 3.14159265358979323846;
// const ll mod = 1000000007;
// const ll mod = 998244353;
// ll mod;
 

 
ll n, m, k, q, l, r, x, y, z;
ll a[1000005];
ll b[1000005];
ll c[1000005];
string s, t;
ll ans = 0;



void solve(int tc) {
  cin >> n >> m;
  if (n % 2) {
    f0r(i, m) cout << i + 1 << " " << n - i << '\n';
  } else {
    ll v = 1 - (n / 2) % 2;
    f0r(i, m / 2 + v) cout << i + 1 << " " << n - i << '\n';
    f1r(i, m / 2 + v, m) cout << i + 1 << " " << n - i - 1 << '\n';
  }
}
 
int main() {
  send help
 
  // usaco("file");
 
  int tc;
  tc = 1;
  // cin >> tc;
  for (int t = 0; t < tc; t++) solve(t);
}