/** gnu specific **/
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
/** contains everything I need in std **/
#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(S) ((int)S.size())
#define FOR(i, n) for(int i = 0; i < n; ++i)
#define RFOR(i, n) for(int i = n-1; i >= 0; --i)
#define output_vec(vec) { FOR(i_, sz(vec)) cout << vec[i_] << ' '; cout << '\n'; }
#define x first
#define y second
#define pb push_back
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<bool> vb;
typedef short si;
typedef unsigned long long ull;
typedef long double LD;
typedef pair<ull, ull> pull;
using namespace __gnu_pbds;
typedef tree<ll, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
using namespace std;
#ifdef ONPC
mt19937 rnd(228);
#else
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif

#define MAX_N 2020
vector<pll> p;
int n;
double d;
vector<int> G[MAX_N];

double dist(const pll& e1, const pll& e2)
{
	return sqrt(pow(e1.x-e2.x, 2) + pow(e1.y - e2.y, 2));
}

void dfs(int v, vector<int>& vis, int marker)
{
	vis[v] = marker;
	for(auto to: G[v])
	{
		if(vis[to] == 0)
		{
			dfs(to, vis, marker);
		}
	}
}

int solve()
{
	if(!(cin >> n >> d))
		return 1;
	p = vector<pll>(n);
	FOR(i, n) cin >> p[i].x >> p[i].y;
	FOR(i, n)
	{
		FOR(j, i)
		{
			if(dist(p[i], p[j]) <= d)
			{
				G[i].pb(j);
				G[j].pb(i);
			}
		}
	}
	vector<int> vis(n, 0);
	int marker = 1;
	FOR(i, n)
	{
		if(vis[i] == 0)
		{
			dfs(i, vis, marker);
			++marker;
		}
	}
	vector<int> S(n+4, 0);
	FOR(i, n)
	{
		S[vis[i]] += 1;
	}
	vector<int> res(n);
	iota(all(res), 0);
	sort(all(res), [&](int e1, int e2) -> bool {
		return S[vis[e1]] > S[vis[e2]];
	});
	FOR(i, n)
		cout << res[i]+1 << ' ';
	cout << '\n';
    return 0;
}

int32_t main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int TET = 1e9;
    //cin >> TET;
    for (int i = 1; i <= TET; i++)
    {
        if (solve())
        {
            break;
        }
#ifdef ONPC
        cout << "__________________________" << endl;
#endif
    }
#ifdef ONPC
    cerr << endl << "finished in " << clock() * 1.0 / CLOCKS_PER_SEC << " sec" << endl;
#endif
}