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 with AC Library v1.1) (judge verdict: Accepted)

#include <atcoder/dsu>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;

vector<int> solve(ll N, vector<int>& X, vector<int>& Y) {
    vector<pii> P(N);
    for ( int i = 0; i < N; i++ ) {
        P[i] = pii(X[i],Y[i]);
    }
    sort(P.begin(), P.end());
    vector<int> yr(N);
    int ym = N;
    for ( int i = 0; i < N; i++ ) {
        int y = P[i].second;
        ym = min(ym, y);
        yr[ym] = max(yr[ym], y);
    }
    for ( int i = 1; i < N; i++ ) {
        yr[i] = max(yr[i],yr[i-1]);
    }
    atcoder::dsu uy(N);
    for ( int i = 0; i < N; i++ ) {
        uy.merge(i,yr[i]);
    }
    vector<int> ans(N);
    for ( int i = 0; i < N; i++ ) {
        ans[i] = uy.size(Y[i]);
    }

    return ans;
}

vector<int> solve_small_1(ll N, vector<int>& X, vector<int>& Y) {
    atcoder::dsu uf(N);
    for ( int i = 0; i < N; i++ ) {
        for ( int j = 0; j < N; j++ ) {
            if ( X[i] < X[j] && Y[i] < Y[j] ) {
                uf.merge(i,j);
            }
        }
    }
    vector<int> ans(N);
    for ( int i = 0; i < N; i++ ) {
        ans[i] = uf.size(i);
    }
    return ans;
}
struct Data {
    int id;
    ll N;
    vector<int> X;
    vector<int> Y;
};
void test() {
    mt19937 mt(0);
    //uniform_int_distribution<ll> rnd_v(0, n-1);

    int n = 100;
    int id = 0;
    vector<Data> ts;
    for ( int i = 0; i < 100; i++ ) {
        vector<int> X(n), Y(n);
        for ( int i = 0; i < n; i++ ) {
            X[i] = i;
            Y[i] = i;
        }
        shuffle(X.begin(), X.end(), mt);
        shuffle(Y.begin(), Y.end(), mt);
        ts.push_back({++id,n,X,Y});
    }
    
    for ( Data& t : ts ) {
        auto ans0 = solve(t.N,t.X,t.Y);
        auto ans1 = solve_small_1(t.N,t.X,t.Y);
        for ( int i = 0; i < t.N; i++ ) {
            if ( ans0[i] != ans1[i] ) {
                cout << t.id << " " << ans0[i] << " " << ans1[i] << "\n";
            }
        }
    }
    cout << "end" << "\n";
}

int main() {
    //test();
    ll N;
    cin >> N;
    vector<int> X(N), Y(N);
    for ( int i = 0; i < N; i++ ) {
        cin >> X[i] >> Y[i];
        X[i]--, Y[i]--;
    }
    auto ans = solve(N,X,Y);
    for ( int i = 0; i < N; i++ ) {
        cout << ans[i] << "\n";
    }    
    return 0;
}