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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

# Binary Indexed Tree (Fenwick Tree)
class BIT:
  def __init__(self, n):
    self.n = n
    self.bit = [0]*(n+1)
    self.el = [0]*(n+1)
  def sum(self, i):
    s = 0
    while i > 0:
      s += self.bit[i]
      i -= i & -i
    return s
  def add(self, i, x):
    # assert i > 0
    self.el[i] += x
    while i <= self.n:
      self.bit[i] += x
      i += i & -i
  def get(self, i, j=None):
    if j is None:
      return self.el[i]
    return self.sum(j) - self.sum(i-1)
  def lower_bound(self,x,p):
    i = p
    w = self.bit[i]
    k = 1<<((self.n).bit_length())
    while k:
      if i+k <= self.n and w + self.bit[i+k] < x:
        w += self.bit[i+k]
        i += k
      k >>= 1
    return i+1

class UnionFind():
  def __init__(self, n):
    self.n = n
    self.parents = [-1] * n

  def find(self, x):
    if self.parents[x] < 0:
      return x
    else:
      self.parents[x] = self.find(self.parents[x])
    return self.parents[x]

  def union(self, x, y):
    x = self.find(x)
    y = self.find(y)

    if x == y:
      return

    if self.parents[x] > self.parents[y]:
      x, y = y, x

    self.parents[x] += self.parents[y]
    self.parents[y] = x

  def same(self, x, y):
    return self.find(x) == self.find(y)

  def roots(self):
    return [i for i, x in enumerate(self.parents) if x < 0]

  def num_roots(self):
    return len([i for i, x in enumerate(self.parents) if x < 0])

  def members(self, x):
    root = self.find(x)
    return [i for i in range(self.n) if self.find(i) == root]

  def num_members(self,x):
    return abs(self.parents[self.find(x)])

  def __str__(self):
    return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

import sys,os,io
input = sys.stdin.readline
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
N = int(input())
A = [list(map(int, input().split())) for i in range(N)]
y_to_i = [0]*(N+1)
for i,(x,y) in enumerate(A):
  y_to_i[y] = i

ans = [1]*N
A.sort()
bit = BIT(N)
uf = UnionFind(N)
cnt = 0
for j,(x,y) in enumerate(A):
  p = bit.sum(y)
  y2 = 0
  for i in range(1,p+1):
    y2 = bit.lower_bound(i,y2)
    if not uf.same(y_to_i[y],y_to_i[y2]):
      uf.union(y_to_i[y],y_to_i[y2])
      cnt += 1
      if cnt>=j:
        break
  bit.add(y,1)
for i in range(N):
  ans[i] = uf.num_members(i)
print(*ans, sep='\n')