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)

#define _USE_MATH_DEFINES

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <atcoder/all>
using namespace std;
using namespace atcoder;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<ll, ll, ll> tl3;

typedef modint998244353 mint;

const int BIG_NUM = 1e9;
const ll INF = 1000000000000000000;
//const ll MOD = 1e9 + 7;
const ll MOD = 998244353;

const ll MAX = 1e9 + 5;

int main() {
	int n;
	cin >> n;

	vector<pii> p(n);
	map<int, pii> m;

	vector<int> yi(n);
	for (int i = 0; i < n; i++) {
		int x, y;
		cin >> x >> y;
		x--;
		y--;
		p[i] = make_pair(x, y);
		m[x] = make_pair(y, i);
		yi[y] = i;
	}
	dsu d(n);
	map<int, pii> m1(m);
	for (int j = n - 1; j >= 0; j--) {

		
		int i = yi[j];
		int xi = p[i].first;
		//cout << xi << endl;
		if (m1.find(xi) == m1.end()) {
			continue;
		}
		for (auto it = m1.begin(); it != m1.find(xi); ) {
			
			
			d.merge(i, it->second.second);
			//cout << i << " "<< it->second.second << endl;
			it =m1.erase(it);
		}
		m1.erase(xi);
	}

	for (int j = 0; j <n ; j++) {


		int i = yi[j];
		int xi = p[i].first;
		//cout << xi << endl;
		if (m.find(xi) == m.end()) {
			continue;
		}
		for (auto it = m.find(xi); it !=m.end() ; ) {


			d.merge(i, it->second.second);
			//cout << i << " "<< it->second.second << endl;
			it = m.erase(it);
		}
		m.erase(xi);
	}
	for (int i = 0; i < n; i++) {
		cout << d.size(i) << endl;
	}

}