#include<bits/stdc++.h>
//#define int long long
#define pb push_back
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define MOD 1000000007
#define inf 1e18
#define fi first
#define se second
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORD(i,a,b) for(int i=a;i>=b;i--)
#define sz(a) ((int)(a).size())
#define endl '\n'
#define pi 3.14159265359
#define TASKNAME "bc"
using namespace std;
template<typename T> bool maximize(T &res, const T &val) { if (res < val){ res = val; return true; }; return false; }
template<typename T> bool minimize(T &res, const T &val) { if (res > val){ res = val; return true; }; return false; }
typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef vector<int> vi;
const int MAXN = 5e3 + 9;
int n;
int x[MAXN], y[MAXN], answer[MAXN];
int dist[5001][5001], g[5001][5001];
void bfs(int s, int dist[]){
queue<int> q;
q.push(s);
dist[s] = 0;
while(!q.empty()){
int u = q.front();
answer[s] += dist[u];
q.pop();
for(int v = 1; v <= n; v++){
if (g[u][v] == 0) continue;
if (minimize(dist[v], dist[u] + 1)){
q.push(v);
}
}
}
}
namespace subtask1{
bool check(){
return n <= 5000;
}
void solve(){
FOR(i, 1, n){
FOR(j, i + 1, n) {
if ((x[i] < x[j] and y[i] < y[j]) or
(x[i] > x[j] and y[i] > y[j])){
g[i][j] = 1;
g[j][i] = 1;
}
}
}
memset(dist, 0x3f, sizeof(dist));
FOR(i, 1, n){
bfs(i, dist[i]);
}
FOR(i, 1, n){
cout << answer[i] << endl;
}
}
}
main()
{
fast;
if (fopen(TASKNAME".inp","r")){
freopen(TASKNAME".inp","r",stdin);
freopen(TASKNAME".out","w",stdout);
}
cin >> n;
FOR(i, 1, n){
cin >>x[i] >> y[i];
}
if (subtask1::check()) return subtask1::solve(), 0;
}
/**
Warning:
Đọc sai đề???
Cận lmao
Code imple thiếu case nào không.
Limit.
**/
