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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

class UnionFind
  attr_accessor(:table)

  def initialize(n)
    @table = Array.new(n, -1)
  end

  def root(x)
    st = []
    tb = @table
    while tb[x] >= 0
      st.push(x)
      x = tb[x]
    end
    
    st.each do |y|
      @table[y] = x
    end
    
    return x
  end
  
  def find(x, y)
    return root(x) == root(y)
  end

  def unite(x, y)
    r1 = root(x)
    r2 = root(y)
    return if r1 == r2
    
    d1 = @table[r1]
    d2 = @table[r2]
    if d1 <= d2
      @table[r2] = r1
      @table[r1] += d2
    else
      @table[r1] = r2
      @table[r2] += d1
    end
  end
end

n = $stdin.readline.to_i
coord = Array.new(n) { $stdin.readline.chomp.split.map(&:to_i) }
coord = coord.map.with_index {|v, i| [v, i].flatten }.sort_by! {|x, _, _| x }
uf = UnionFind.new(n)

0.step(n - 2) do |i|
  (i + 1).step(n - 1) do |j|
    uf.unite(coord[i][2], coord[j][2]) if coord[i][1] < coord[j][1]
  end
end
n.times {|v| puts (- uf.table[uf.root(v)]) }