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<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;
#define int long long
#define double long double
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
//printf("%.12Lf\n",);
int keta(int x){
	rep(0,i,30){
		if(x<10){
			return i+1;
		}
		x=x/10;
	}
}	
int gcd(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return bb;
		}	
		bb=bb%aa;
		if(bb==0){
			return aa;
		}
	}
}	
int lcm(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return x/bb*y;
		}	
		bb=bb%aa;
		if(bb==0){
			return x/aa*y;
		}
	}
}
bool p(int x){
	if(x==1)return false;
	rep(2,i,sqrt(x)+1){
		if(x%i==0&&x!=i){
			return false;
		}
	}	
	return true;
}
int max(int a,int b){
	if(a>=b)return a;
	else return b;
}
string maxst(string s,string t){
	int n=s.size();
	int m=t.size();
	if(n>m)return s;
	else if(n<m)return t;
	else{
		rep(0,i,n){
			if(s[i]>t[i])return s;
			if(s[i]<t[i])return t;
		}
		return s;
	}
}		
int min(int a,int b){
	if(a>=b)return b;
	else return a;
}	
int n2[41];		
int nis[41]; 
int nia[41];
int mody[41];
int nn;
int com(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int bunsi=1,bunbo=1;
	rep(0,i,y)bunsi=(bunsi*(n-i))%mod;
	rep(0,i,y)bunbo=(bunbo*(i+1))%mod;
	mody[0]=bunbo;
	rep(1,i,41){
		bunbo=(bunbo*bunbo)%mod;
		mody[i]=bunbo;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			bunsi=(bunsi*mody[i])%mod;
		}
	}
	return bunsi;
}
int gyakugen(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	mody[0]=y;
	rep(1,i,41){
		y=(y*y)%mod;
		mody[i]=y;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			n=(n*mody[i])%mod;
		}
	}
	return n;	
}	
int yakuwa(int n){
	int sum=0;
	rep(1,i,sqrt(n+1)){
		if(n%i==0)sum+=i+n/i;
		if(i*i==n)sum-=i;
	}
	return sum;
}	
bool integer(double n){
	if(n==long(n))return true;
	else return false;	
}
int poow(int y,int n){	
	if(n==0)return 1;
	n-=1;
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int yy=y;
	mody[0]=yy;
	rep(1,i,41){
		yy=(yy*yy)%mod;
		mody[i]=yy;
	}
	rep(0,i,41)nis[i]=0;
	nn=n;
	for(int i=40;i>=0;i-=1){
		if(nn>=n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	rep(0,i,41){
		if(nis[i]==1){
			y=(y*mody[i])%mod;
		}
	}
	return y;	
}
int minpow(int x,int y){
	int sum=1;
	rep(0,i,y)sum*=x;
	return sum;
}
int ketawa(int x,int sinsuu){
	int sum=0;
	rep(0,i,100)sum+=(x%poow(sinsuu,i+1))/(poow(sinsuu,i));
	return sum;
}	
int sankaku(int a){
	return a*(a+1)/2;
} 
bool search(int x){
	int n;
	int a[n];
	int l=0,r=n;
	while(r-l>=1){
		int i=(l+r)/2;
		if(a[i]==x)return true;
		else if(a[i]<x)l=i+1;
		else r=i;
	}
	return false;
}
/*using Graph=vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
    depth[v] = d;
    oya[v]=p;
    for (auto nv : G[v]) {
       if (nv == p) continue; // nv が親 p だったらダメ
       dfs(G, nv, v, d+1); // d を 1 増やして子ノードへ
    }
}*/
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
    memo[h][w] = 'x';

    // 八方向を探索
    for (int dh = -1; dh <= 1; ++dh) {
        for (int dw = -1; dw <= 1; ++dw) {
        	if(abs(0-dh)+abs(0-dw)==2)continue;
            int nh = h + dh, nw = w + dw;

            // 場外アウトしたり、0 だったりはスルー
            if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
            if (memo[nh][nw] == 'x') continue;

            // 再帰的に探索
            dfs(nh, nw);
        }
    }
}*/
int XOR(int a,int b){
	int ni=1;
	rep(0,i,41){
		n2[i]=ni;
		ni*=2;
	}
	rep(0,i,41)nis[i]=0;
	for(int i=40;i>=0;i-=1){
		if(a>=n2[i]){
			nis[i]++;
			a-=n2[i];
		}
		if(b>=n2[i]){
			nis[i]++;
			b-=n2[i];
		}	
	}
	int sum=0;
	rep(0,i,41)sum+=(nis[i]%2*n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
bool a[114514];
signed main(){
	ic(n) ic(m) 
	if(n%2==0){
		n-=1;
	}
	if(m*2+1==n){
		rep(0,i,m)cout<<i+1<<" "<<n-i-1<<endl;
		return 0;
	}	
	int k=(n-1)/2;
	if(k%2==0)k-=1;
	rep(0,i,k)a[i]=false;
	if(m%2==1){
		a[(k-1)/2]=true;
		m-=1;
	}	
	a[0]=true;
	a[k-1]=true;
	rep(0,i,m/2){
		a[(k-1)/2-i-1]=true;
		a[(k+1)/2+i]=true;
	}
	rep(0,i,k){
		if(a[i])cout<<i+1<<" "<<n-i-1<<endl;
	}
}