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: Wrong Answer)

#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
typedef long long ll;
int main() {
	int N;
	cin >> N;
	vector<int> X(N), Y(N);
	vector<int> a(N), b(N);
	for (int i = 0; i < N; ++ i) {
		cin >> X[i] >> Y[i];
		-- X[i]; -- Y[i];
		a[X[i]] = Y[i];
		b[Y[i]] = X[i];
	}
	dsu D(N);
	{
		set<int> hoge;
		for (int i = 0; i < N; ++ i) {
			for (auto it = hoge.begin(); it != hoge.end(); ) {
				if (*it < a[i]) {
					D.merge(b[*it], i);
					it = hoge.erase(it);
				} else {
					break;
				}
			}
			hoge.insert(a[i]);
		}
	}
	{
		set<int, greater<int>> hoge;
		for (int i = N-1; i >= 0; -- i) {
			for (auto it = hoge.begin(); it != hoge.end(); ) {
				if (*it > a[i]) {
					D.merge(b[*it], i);
					it = hoge.erase(it);
				} else {
					break;
				}
			}
			hoge.insert(a[i]);
		}
	}
	vector<int> res(N);
	for (auto g : D.groups()) {
		int n = g.size();
		for (int x : g) res[x] = n;
	}
	for (int i = 0; i < N; ++ i) {
		cout << res[X[i]] << endl;
	}
}