///by Saton.
#include<bits/stdc++.h>
#define PI acos(-1)
#define fi first
#define se second 
#define sz(a) ((int)a.size())
#define all(a) a.begin(), a.end()
#define LL long long
#define ll __int128
#define DD double double
#define LD long double
#define rep(i,a,b) for(LL i = (a);i <= (b);i ++)
#define lep(i,a,b) for(LL i = (a);i >= (b);i --)
#define FLUSH fflush(stdout)
using namespace std;
const int N = 2e5 + 10,mod =  1e9+7,P = 131;
const LL inf = 1e9+10,INF = 1e18+10;
typedef pair<int,int> PII;
typedef pair<LL,int> PLI;
typedef pair<LL,LL> PLL;
//__builtin_popcountll()计算整数二进制表示中1的个数
// mt19937_64 rnd(random_device{}());
// uniform_int_distribution<LL> dist(0, LLONG_MAX);//dist(rnd)生成随机数
LL n,m,k;
string str;
int dx[4] = {-1,1,0,0},dy[4] = {0,0,-1,1};
// __int128//最大可用整数类型
// struct Node {
//     int l,r;
//     bool operator<(const Node &u) const {
//         return l < u.l;
//     }
// }node[N];
 
LL qmi(LL a,LL b,LL p) {LL res = 1;while(b) {if(b&1) res = (LL)res*a%p;a = (LL)a*a%p;b >>= 1;}return res%p;}
LL gcd(LL a,LL b) {return b ? gcd(b,a%b) : a;}
LL lcm(LL a,LL b) {return (LL)a*b / gcd(a,b);}
LL lowbit(LL x) {return x & -x;} 

void solve() {
    cin >> n >> k;
    vector<PII> ans;
    map<PII,int> mp;
    rep(i,1,n-1) {
        ans.push_back({i,i});
        mp[{i,i}] = 1;
        k --;
        ans.push_back({i,i+1});
        mp[{i,i+1}] = 1;
        k --;
    }
    ans.push_back({n,n});
    mp[{n,n}] = 1;
    k --;
    ans.push_back({n,1});
    mp[{n,1}] = 1;
    k --;    
    for(int i = 1;i<=n && k;i ++) {
        for(int j = 1;j<=n && k;j ++) {
            if(!mp[{i,j}]) {
                ans.push_back({i,j});
                mp[{i,j}] = 1;
                k --;
            }
        }
    }
    
    for(auto [u,v] : ans) cout << u << " " << v << '\n';
} 

int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    // int T;
    // cin >> T;
    // while(T --) { 
    //     solve();
    // }
    solve();
       
    return 0;
}  
/*   /\_/\
*   (= ._.)
*   / >  \>
*/