Problem:
Score : 300 points

Problem Statement

There are N cities on a 2D plane. The coordinate of the i-th city is (x_i, y_i). Here (x_1, x_2, \dots, x_N) and (y_1, y_2, \dots, y_N) are both permuations of (1, 2, \dots, N).

For each k = 1,2,\dots,N, find the answer to the following question:

Rng is in City k.
Rng can perform the following move arbitrarily many times:

move to another city that has a smaller x-coordinate and a smaller y-coordinate, or a larger x-coordinate and a larger y-coordinate, than the city he is currently in.

How many cities (including City k) are reachable from City k?

Constraints

1 \leq N \leq 200,000

(x_1, x_2, \dots, x_N) is a permutation of (1, 2, \dots, N).

(y_1, y_2, \dots, y_N) is a permutation of (1, 2, \dots, N).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print N lines. In i-th line print the answer to the question when k = i.

Sample Input 1

4
1 4
2 3
3 1
4 2

Sample Output 1

1
1
2
2

Rng can reach City 4 from City 3, or conversely City 3 from City 4.

Sample Input 2

7
6 4
4 3
3 5
7 1
2 7
5 2
1 6

Sample Output 2

3
3
1
1
2
3
2

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define task "sort"
#define pll pair<ll, ll>
#define pii pair<ll, pll>
#define fi first
#define se second
#define ull unsigned long long
using namespace std;
const ll mod = 1e9;
const ll N = 2e5 + 5;

vector<ll> adj[N];
vector<ll> kq;

priority_queue< pll, vector<pll>, greater<pll> > pr, prn;
ll n, m, t, k, b[N], lab[N], res[N], ans, tong;
ll findp(ll u)
{
    return lab[u] < 0 ? u : lab[u] = findp(lab[u]);
}
void hop(ll u, ll v)
{
    if(lab[u] > lab[v])swap(u, v);
    lab[u] += lab[v];
    lab[v] = u;
}
struct dang
{
    ll l, r, pos;
} a[N];
bool cmp(const dang& X, const dang& Y)
{
    return X.l > Y.l;
}
bool cmt(const dang& X, const dang& Y)
{
    return X.l < Y.l;
}
void dfs(ll u)
{
    b[u] = 1;
    for(ll v : adj[u])
    {
        if(lab[v] == 0)
        {
            lab[v] = lab[u];
            dfs(v);
            b[u] += b[v];
        }
    }
}
void sol()
{
    cin >> n;
    for(int i = 1; i <= n; i ++)
    {
        cin >> a[i].l >> a[i].r;
        a[i].pos = i;
        b[a[i].l] = i;
        lab[i] = -1;
    }
    sort(a+1, a+1+n, cmt);
    set<ll> ds;
    for(int i = 1; i <= n; i ++)
    {

        auto it = ds.lower_bound(-a[i].r);
        if(it != ds.end())
        {
            //cout << b[-*it] <<" "<<a[i].pos << '\n';
            ll u = findp(b[-*it]), v = findp(a[i].pos);
            if(u != v)hop(u, v);
        }
        else ds.insert(-a[i].r);
        b[a[i].r] = a[i].pos;

    }
    ds.clear();
    sort(a+1, a+1+n, cmp);
    for(int i = 1; i <= n; i ++)
    {

        auto it = ds.lower_bound(a[i].r);
        if(it != ds.end())
        {
            //cout << b[-*it] <<" "<<a[i].pos << '\n';
            ll u = findp(b[*it]), v = findp(a[i].pos);
            if(u != v)hop(u, v);
        }
        else ds.insert(a[i].r);
        b[a[i].r] = a[i].pos;

    }
    for(int i = 1; i <= n; i ++)
    {
        ll u = findp(i);
        cout << - lab[u] << '\n';
    }

}
int main()
{
    if(fopen(task".in", "r")){
       freopen(task".in", "r", stdin);
       freopen(task".out", "w", stdout);
    }
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    sol();
}