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: Time Limit Exceeded)

#include <atcoder/all>
#include <bits/stdc++.h>
using namespace atcoder;
using namespace std;

typedef pair<int, int> ii;
typedef pair<ii, int> iii;

int main()
{
	int n;
	scanf("%d", &n);
	vector<iii> p(n);
	vector<int> rev(n);
	dsu d(n);
	for (int i = 0; i < n; ++i) scanf("%d%d", &p[i].first.first, &p[i].first.second), p[i].second = i, rev[p[i].first.second-1] = i;
	sort(p.begin(), p.end());
	set<int> r;
	int prev = -1;
	for (int i = 0; i < n; ++i) r.insert(lower_bound(r.begin(), r.end(), p[i].first.second), p[i].first.second);
	for (int i = 0; i < n; ++i) {
		int pt = p[i].first.second;
		r.erase(pt);
		auto it = r.lower_bound(pt);
		if (*it == prev) {
			d.merge(p[i].second, p[i-1].second);
		}
		else {
			for (auto end = r.end(); it != end; ++it) {
				d.merge(p[i].second, rev[(*it)-1]);
			}
			prev = it != r.end() ? *it : -1;
		}
	}
	for (int i = 0; i < n; ++i) printf("%d\n", d.size(i));
}