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;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> ii;
typedef tuple<ll, ll, ll> iii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
#define REP(i,n) for (ll i = 0; i < n; ++i)
#define REPR(i,n) for (ll i = n-1; i >= 0; --i)
#define FOR(i,m,n) for (ll i = m; i < n; ++i)
#define FORR(i,m,n) for (ll i = n-1; i >= m; --i)
#define FORE(x,xs) for (const auto& x : xs)
#define FORI(i,v) for (auto i = v.begin(); i != v.end(); i++)
#define ALL(v) v.begin(), v.end()
#define CHMIN(x,y) x = min(x, y)
#define CHMAX(x,y) x = max(x, y)
#define YES(b) cout << ((b) ? "YES" : "NO") << endl
#define Yes(b) cout << ((b) ? "Yes" : "No") << endl
#define DOUBLE(d) cout << fixed << setprecision(15) << (d) << endl

int N, M;

void solve() {
    // 差が奇数
    int a = 1;
    int b = (M+1)/2*2;
    REP (i, (M+1)/2) {
        cout << a << " " << b << endl;
        a++; b--;
    }

    // 差が偶数
    a = (M+1)/2*2+1;
    b = 2*M+1;
    REP (i, M/2) {
        cout << a << " " << b << endl;
        a++; b--;
    }
}

int main() {
    cin >> N >> M;
    solve();
}