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 Python (3.8.2) (judge verdict: Accepted)

from heapq import heappush, heappop

class UnionFind:
    def __init__(self, size):
        self.data = [-1] * size
    def find(self, x):
        if self.data[x] < 0:
            return x
        else:
            self.data[x] = self.find(self.data[x])
            return self.data[x]
    def union(self, x, y):
        x, y = self.find(x), self.find(y)
        if x != y:
            if self.data[y] < self.data[x]:
                x, y = y, x
            self.data[x] += self.data[y]
            self.data[y] = x
        return (x != y)
    def same(self, x, y):
        return (self.find(x) == self.find(y))
    def size(self, x):
        return -self.data[self.find(x)]

N, *XY = map(int, open(0).read().split())
YtoI = {y : i for i, y in enumerate(XY[1::2])}

uf = UnionFind(N)
Q = []
for x, y in sorted(zip(*[iter(XY)] * 2)):
    heappush(Q, y)
    mi = Q[0]
    while Q and Q[0] <= y:
        uf.union(YtoI[y], YtoI[heappop(Q)])
    heappush(Q, mi)

for i in range(N):
    print(uf.size(i))