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: Wrong Answer)

import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


class UnionFindTree:
    def __init__(self, n):
        self.n = n
        self.parent = list(range(n))
        self.size = [1] * n

    def root(self, i):
        inter = set()
        while self.parent[i]!=i:
            inter.add(i)
            i = self.parent[i]
        r = i
        for i in inter:
            self.parent[i] = r
        return r

    def connect(self, i, j):
        ri = self.root(i)
        rj = self.root(j)
        if ri==rj:
            return
        if self.size[ri]<self.size[rj]:
            self.parent[ri] = rj
            self.size[rj] += self.size[ri]
        else:
            self.parent[rj] = ri
            self.size[ri] += self.size[rj]
### BIT binary
def init(bit, values):
    for i,v in enumerate(values):
        add(bit,i+1,v)
#a1 ~ aiまでの和 O(logn)
def query(bit,i):
    res = 0
    while i > 0:
        res += bit[i]
        i -= i&(-i)
    return res

#ai += x(logN)
def add(bit,i,x):
    if i==0:
        raise RuntimeError
    while i <= len(bit)-1:
        bit[i] += x
        i += i&(-i)
    return

def index(bit, v):
    """a1,...,aiの和がv以上になる最小のindexを求める
    存在しないとき配列サイズを返す
    """
    i = 0
    n = len(bit)-1
    k = 1<<((n-1).bit_length())
    while k>0:
        if i+k<n+1 and bit[i+k]<v:
            v -= bit[i+k]
            i += k
        k //= 2
    if v==1:
        return i+1
    else:
        return n+1
n = int(input())
xy = []
dx = {}
dy = {}
for i in range(n):
    x,y = map(lambda x: int(x)-1, input().split())
    xy.append((i,x,y))
    dx[x] = i
    dy[y] = i

uf = UnionFindTree(n)
xy.sort(key=lambda item: item[1])

bit = [0]*(n+1)
for i,x,y in xy:
    num = query(bit, y+1)
    if num>0:
        ii = dy[index(bit, num)-1]
        uf.connect(i,ii)
    add(bit, y+1, 1)
bit = [0]*(n+1)
c = 0
for i,x,y in reversed(xy):
    num = query(bit, y+1)
    if c-num>0:
        ii = dy[index(bit, num+1)-1]
        uf.connect(i,ii)
    add(bit, y+1, 1)
    c += 1

xy.sort(key=lambda item: item[2])
bit = [0]*(n+1)
for i,x,y in xy:
    num = query(bit, x+1)
    if num>0:
        ii = dx[index(bit, num)-1]
        uf.connect(i,ii)
    add(bit, x+1, 1)
bit = [0]*(n+1)
c = 0
for i,x,y in reversed(xy):
    num = query(bit, x+1)
    if c-num>0:
        ii = dx[index(bit, num+1)-1]
        uf.connect(i,ii)
    add(bit, x+1, 1)
    c += 1

ans = [None]*n
for i in range(n):
    ans[i] = uf.size[uf.parent[i]]
write("\n".join(map(str, ans)))