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>
#define repr(i,from,to) for(int (i)=(from);(i)<(to);(i)++)
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
// const bool debug=true;
const bool debug=false;
#define DEBUG if(debug==true)
#define all(x) begin((x)), end((x))
#define vprint(x) for(auto a:(x)) cout << x << endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 1000000007;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) { a = b; return true;}
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) { a = b; return true;}
    return false;
}


int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,m;
    cin>>n>>m;
    if (n%2==1) {
        ll sm = n/2, lg = sm+1;
        rep(i,m) {
            cout << sm << " " << lg << endl;
            sm--;lg++;
        }
    }
    else {
        ll ue = n / 2, sita = ue;
        ll lg,sm, cnt;
        if (n % 4 == 0) {
            sm = sita + 2; lg = n;
            cnt = ue / 2 - 1;
        }
        else {
            ue -= 1;
            sm = sita + 1; lg = n;
            cnt = ue / 2;
        }
        while(m>0) {
            if (cnt == 0) {
                if (n % 4 == 0) {
                    sm = 1;
                    lg = sita;
                }
                else {
                    sm = 1;
                    lg = sita-1;
                }
            }
            cout << sm << " " << lg << endl;
            sm++;lg--;cnt--;
            m--;
        }
    }
    return 0;
}