#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define MASK(x) (1LL << (x))
#define BIT(x, i) (((x) >> (i)) & 1)
#define SZ(x) ((int) (x).size())
#define ALL(a) (a).begin(), (a).end()
#define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; ++i)
#define FORD(i, a, b) for (int i = (a), _b = (b); i >= _b; --i)
#define REP(i, n) for (int i = 0, _n = (n); i < _n; ++i)
#define debug(x) cout << #x << " = " << (x) << endl;
#define fi first
#define se second
#define left __left
#define right __right
#define prev __prev
#define next __next
template <class X, class Y>
bool maximize(X &x, Y y) {
if (x < y) return x = y, true;
else return false;
}
template <class X, class Y>
bool minimize(X &x, Y y) {
if (x > y) return x = y, true;
else return false;
}
template <class T>
void remove_dup(T &v) {
sort(ALL(v));
v.resize(unique(ALL(v)) - v.begin());
}
int POS(int x, const vector <int> &v) {
return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
const int inf = (int) 1e9 + 7;
#define MAX_N 300'300
int n;
int del[MAX_N + 2];
struct Circle {
int x, y, r, id;
bool operator < (const Circle &other) const {
if (r != other.r) return (r > other.r);
return (id < other.id);
}
friend istream& operator >> (istream &in, Circle &c) {
in >> c.x >> c.y >> c.r;
return in;
}
} cir[MAX_N + 2];
ll sqr(ll x) {
return 1LL * x * x;
}
namespace subtask1 {
bool check() {
return (n <= 5000);
}
void solve() {
sort(cir + 1, cir + 1 + n);
memset(del, 0, sizeof(del));
FOR(i, 1, n) {
int curID = cir[i].id;
if (del[curID]) continue;
del[curID] = curID;
FOR(j, i + 1, n) {
if (del[cir[j].id]) continue;
if (sqr(cir[i].x - cir[j].x) + sqr(cir[i].y - cir[j].y) <= sqr(cir[i].r + cir[j].r)) {
del[cir[j].id] = curID;
}
}
}
FOR(i, 1, n) cout << del[i] << " ";
}
};
const int TREE_SIZE = MAX_N * 2;
struct Segtree {
int seg[TREE_SIZE * 4 + 2], lazy[TREE_SIZE * 4 + 2];
int n;
void initTree(int _n) {
n = _n;
FOR(i, 0, 4 * _n + 7) {
seg[i] = inf;
lazy[i] = inf;
}
}
void down(int id, int l, int r) {
if (lazy[id] == inf || l >= r) return;
int &v = lazy[id];
FOR(j, 2 * id, 2 * id + 1) {
minimize(seg[j], v);
minimize(lazy[j], v);
}
v = inf;
}
void update(int id, int l, int r, int u, int v, int val) {
if (l > v || u > r) return;
if (u <= l && r <= v) {
minimize(seg[id], val);
minimize(lazy[id], val);
return;
}
down(id, l, r);
int g = (l + r) >> 1;
update(id << 1, l, g, u, v, val);
update(id << 1 | 1, g + 1, r, u, v, val);
seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
}
int get(int id, int l, int r, int u, int v) {
if (l > v || u > r) return inf;
if (u <= l && r <= v) return seg[id];
down(id, l, r);
int g = (l + r) >> 1;
return min(get(id << 1, l, g, u, v), get(id << 1 | 1, g + 1, r, u, v));
}
void update(int u, int v, int val) {
update(1, 1, n, u, v, val);
}
int get(int u, int v) {
if (u > v) return inf;
return get(1, 1, n, u, v);
}
} st;
namespace subtask2 {
bool check() {
FOR(i, 1, n) if (cir[i].y != 0) return false;
return true;
}
void solve() {
sort(cir + 1, cir + 1 + n);
vector <int> compressed;
FOR(i, 1, n) {
compressed.push_back(cir[i].x - cir[i].r);
compressed.push_back(cir[i].x + cir[i].r);
}
remove_dup(compressed);
int lim = SZ(compressed) + 2;
st.initTree(lim);
memset(del, 0, sizeof(del));
FOR(i, 1, n) {
int curID = cir[i].id;
int L = POS(cir[i].x - cir[i].r, compressed), R = POS(cir[i].x + cir[i].r, compressed);
int get_val = st.get(L, R);
if (get_val != inf) del[curID] = cir[get_val].id;
else {
// assert(get_val == inf);
del[curID] = curID;
st.update(L, R, i);
}
}
FOR(i, 1, n) cout << del[i] << " ";
}
};
const int dx[] = {-2, -1, 0, 1, 2};
const int dy[] = {-2, -1, 0, 1, 2};
namespace subtask3 {
bool check() {
FOR(i, 2, n) if (cir[i].r != cir[1].r) return false;
return true;
}
// ko can xoa vi tong lan duyet ko qua lon
/// sum visit all circle <= 50 * n;
pair <int, int> nen[MAX_N + 2];
struct Data {
pair <int, int> dir;
int id;
bool operator < (const Data &other) const {
if (dir != other.dir) return (dir < other.dir);
return (id < other.id);
}
};
void solve() {
/// compressed into board with each cell have size R x R;
FOR(i, 1, n) nen[i] = {cir[i].x / cir[i].r, cir[i].y / cir[i].r};
vector <Data> v;
FOR(i, 1, n) v.push_back({nen[i], i});
sort(ALL(v));
FOR(i, 1, n) {
if (del[i]) continue;
/// kc toi da de 2 o giao nhau;
FOR(nx, nen[i].fi -2, nen[i].fi + 2) FOR(ny, nen[i].se -2, nen[i].se + 2) {
pair <int, int> curDir = {nx, ny};
int l = 0, r = SZ(v) - 1, g, pos = -1;
while (l <= r) {
g = (l + r) >> 1;
if (v[pos].dir < curDir) l = g + 1;
else pos = g, r = g - 1;
}
if (pos == -1) continue;
while (pos < SZ(v)) {
pair <int, int> tmp = v[pos].dir;
if (tmp != curDir) break;
int id = v[pos].id;
if (!del[id] && sqr(cir[i].x - cir[id].x) + sqr(cir[i].y - cir[id].y) <= sqr(cir[i].r + cir[id].r)) {
del[id] = i;
}
++pos;
}
}
}
FOR(i, 1, n) cout << del[i] << " ";
}
};
namespace subtask4 {
void solve() {
sort(cir + 1, cir + 1 + n);
}
};
int main() {
ios_base::sync_with_stdio(false);cin.tie(nullptr);
cin >> n;
FOR(i, 1, n) cin >> cir[i], cir[i].id = i;
if (subtask1 :: check()) return subtask1 :: solve(), 0;
if (subtask2 :: check()) return subtask2 :: solve(), 0;
if (subtask3 :: check()) return subtask3 :: solve(), 0;
assert(false);
return 0;
}
