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 <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <set>
#include <iomanip>
#include <queue>
#include <string>
#include <map>
#include <fstream>
#include <cassert>
#include <stack>
#include <climits>
#include <array>
#include <unordered_set>
#include <unordered_map>
#include <memory>
#include <functional>
#include <cfloat>
#include <numeric>
#include <atcoder/dsu>
struct Coordinate {
	int x, y;
};
int main() {
	int n; std::cin >> n;
	std::vector<Coordinate> coordinate(n);
	for (auto& c : coordinate) {
		std::cin >> c.x >> c.y;
	}
	atcoder::dsu uft(n);
	std::stack<std::pair<int, int>> stack;
	std::vector<int> indices(n); std::iota(indices.begin(), indices.end(), 0);
	std::sort(indices.begin(), indices.end(), [&](const int i, const int j) {return coordinate[i].x < coordinate[j].x; });
	for (const auto i : indices) {
		while (!stack.empty() && stack.top().second < coordinate[i].y) {
			uft.merge(i, stack.top().first);
			stack.pop();
		}
		stack.emplace(i, coordinate[i].y);
	}
	std::reverse(indices.begin(), indices.end());
	while (!stack.empty()) stack.pop();
	for (const auto i : indices) {
		while (!stack.empty() && stack.top().second > coordinate[i].y) {
			uft.merge(i, stack.top().first);
			stack.pop();
		}
		stack.emplace(i, coordinate[i].y);
	}
	for (auto i = 0; i < n; ++i) {
		std::cout << uft.size(i) << '\n';
	}
}