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 <bits/stdc++.h>
#include <atcoder/dsu>
#include <cstdio>
using namespace std;
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define rep3(i, s, n) for (int i = (s); i <= (int)(n); i++)

int main() {
	int N;
	int Q;
	int q_ll,q_rr,x;
	int r;
	int ret;
	int retl,p;

	cin >> N;
	vector<int> data(N+1);//良い位置には1、悪い位置は0を代入
    dsu d(N+1);
	vector<int> x_y(N+1);//大きい方
	vector<vector<int>> qj(N+1);//問題の番号管理
	vector<int> x_l(N+1);//小さい方
	vector<int> qi(N+1);//問題の番号管理
	vector<int> ans(N+1);//答え

	rep3(i, 1, N) data.at(i)=0;
	rep3(i, 1, N) {
		cin >> q_ll >> q_rr;
		x_l.at(q_ll)=i;
		x_y.at(q_ll)=q_rr;
		qi.at(q_rr)=i;
	}
	rep3(i, 1, N) {
		x=x_y.at(i);
		p=x;
      for (int i = x-1; i>0; i--){
			if(data.at(i)>0){
				data.at(i)=0;
				d.merge(i, p);
 				p=i;
			}
		} 
		data.at(p)=1;
	}
	rep3(i, 1, N) ans.at(x_l.at(i))=d.size(x_y.at(i));
	rep3(i, 1, N) cout << ans.at(i) << endl;
	return 0;
}