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/all>
#include <bits/stdc++.h>

using namespace std;
using namespace atcoder;
typedef long long ll;
typedef long double Double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

// chmax, chmin
template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}


template<typename T>
class FenwickTree {
    const int n;
    std::vector<T> data;

public:
    FenwickTree(int count) : n(count), data(count, 0) { ; }

    void add(int pos, const T &value) {
        assert(0 <= pos && pos < n);
        for (int i = pos; i < n; i |= i + 1) data[i] += value;
    }

    // 区間 [0, pos) 番目の範囲の和を求める．(pos = 0 のときは 0 を返す．)
    T sum(int pos) const {
        assert(0 <= pos && pos <= n);
        T res = 0;
        for (int i = pos - 1; i >= 0; i = (i & (i + 1)) - 1) {
            res += data[i];
        }
        return res;
    }

    // 区間 [l, r) 番目の範囲の和を求める．(l = r のときは 0 を返す．)
    T sum(int l, int r) const {
        assert(0 <= l && l <= r && r <= n);
        return sum(r) + (-sum(l));
    }

    using value_type = T;
    using update_type = T;
};


void solve(long long N, std::vector<long long> x, std::vector<long long> y) {
    auto d = dsu(N);


    vector<pair<pair<int, int>, int> > ps;
    for (int i = 0; i < N; i++) {
        x[i]--;
        y[i]--;
        ps.push_back({{x[i], y[i]}, i});
    }

    vector<int> res(N);
    {


        sort(ps.begin(), ps.end());

        vector<pair<pair<int, int>, int>> Q;
        Q.push_back(ps[0]);

        for (int i = 1; i < ps.size(); i++) {
            auto p = ps[i];
            vector<pair<pair<int, int>, int>> targets;
            while (Q.size() && Q.back().first.second < p.first.second) {
                targets.push_back(Q.back());
                Q.pop_back();
            }
            if (targets.size() >= 1) {
                Q.push_back(targets.front());
                for (auto t : targets) {
                    d.merge(t.second, p.second);
                }
            } else {
                Q.push_back(p);
            }
        }
    }
    {
        sort(ps.rbegin(), ps.rend());

        vector<pair<pair<int, int>, int>> Q;
        Q.push_back(ps[0]);

        for (int i = 1; i < ps.size(); i++) {
            auto p = ps[i];
            vector<pair<pair<int, int>, int>> targets;
            while (Q.size() && Q.back().first.second > p.first.second) {
                targets.push_back(Q.back());
                Q.pop_back();
            }
            if (targets.size() >= 1) {
                Q.push_back(targets.front());
                for (auto t : targets) {
                    d.merge(t.second, p.second);
                }
            } else {
                Q.push_back(p);
            }
        }
    }
    for (int i = 0; i < N; i++) {
        cout << d.size(i) << endl;
    }


}

int main() {
    long long N;
    scanf("%lld", &N);
    std::vector<long long> x(N);
    std::vector<long long> y(N);
    for (int i = 0; i < N; i++) {
        scanf("%lld", &x[i]);
        scanf("%lld", &y[i]);
    }
    solve(N, std::move(x), std::move(y));
    return 0;
}