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) (judge verdict: Accepted)

#include <bits/stdc++.h>

// #include <atcoder/>

using namespace std;

class DisjointSet {
 private:
  class Node {
    friend DisjointSet;

    std::size_t parent;
    std::size_t size;

    Node(const std::size_t parent, const std::size_t size) : parent(parent), size(size) {}
  };

  std::vector<Node> nodes;

 public:
  explicit DisjointSet(const std::size_t n) : nodes(n, Node(n, 1)) {}

  std::size_t size() const { return nodes.size(); }

  std::size_t size(const std::size_t x) { return nodes[root(x)].size; }

  std::size_t root(const std::size_t x) {
    if (nodes[x].parent == size()) return x;
    return nodes[x].parent = root(nodes[x].parent);
  }

  bool is_same(const std::size_t x, const std::size_t y) { return root(x) == root(y); }

  bool unite(const std::size_t x, const std::size_t y) {
    std::size_t rx = root(x), ry = root(y);
    if (rx == ry) return false;
    if (nodes[rx].size < nodes[ry].size) std::swap(rx, ry);
    nodes[rx].size += nodes[ry].size;
    nodes[ry].parent = rx;
    return true;
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<int> X(N), Y(N);
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i];
  }
  DisjointSet ds(N);
  {
    vector<int> idx(N);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(), [&](int a, int b) { return X[a] < X[b]; });
    int pos = 0;
    for (int i = 0; i < N; i++) {
      int a = idx[pos], b = idx[i];
      if (Y[a] > Y[b]) pos = i;
      if (X[a] < X[b] && Y[a] < Y[b]) ds.unite(a, b);
    }
    reverse(idx.begin(), idx.end());
    pos = 0;
    for (int i = 0; i < N; i++) {
      int a = idx[pos], b = idx[i];
      if (Y[a] < Y[b]) pos = i;
      if (X[a] > X[b] && Y[a] > Y[b]) ds.unite(a, b);
    }
  }
  {
    vector<int> idx(N);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(), [&](int a, int b) { return Y[a] < Y[b]; });
    int pos = 0;
    for (int i = 0; i < N; i++) {
      int a = idx[pos], b = idx[i];
      if (X[a] > X[b]) pos = i;
      if (X[a] < X[b] && Y[a] < Y[b]) ds.unite(a, b);
    }
    reverse(idx.begin(), idx.end());
    pos = 0;
    for (int i = 0; i < N; i++) {
      int a = idx[pos], b = idx[i];
      if (X[a] < X[b]) pos = i;
      if (X[a] > X[b] && Y[a] > Y[b]) ds.unite(a, b);
    }
  }
  for (int i = 0; i < N; i++) {
    cout << ds.size(i) << '\n';
  }
  return 0;
}