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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;

// input
int N;
vector<int> x, y;


void input() {
    cin >> N;
    x = y = vector<int>(N);
    rep(i, N) cin >> x[i] >> y[i];
}


struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int n) : par(n) { //最初は全てが根であるとして初期化
        for(int i = 0; i < n; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main() {
    input();
    UnionFind uf(N);
    set<P> ylist;
    vector<T> w;
    rep(i, N) {
        ylist.insert(P(y[i], i));
        w.push_back(tuple(x[i], y[i], i));
    }
    sort(w.begin(), w.end());
    set<int> s;
    for (T t : w) {
        int x_, y_, i;
        tie(x_, y_, i) = t;
        ylist.erase(P(y_, i));
        if (s.count(i)) continue;
        s.insert(i);
        auto itr = ylist.upper_bound(P(y_, i));
        while (itr != ylist.end()) {
            int j = itr->second;
            uf.unite(i, j);
            s.insert(j);
            itr++;
        }
    }
    map<int, int> m;
    rep(i, N) m[uf.root(i)]++;
    rep(i, N) cout << m[uf.root(i)] << endl;
}