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)

class Union_Find():
    def __init__(self,N):
        """0,1,...,n-1を要素として初期化する.

        n:要素数
        """
        self.n=N
        self.parents=[-1]*N
        self.rank=[0]*N

    def find(self, x):
        """要素xの属している族を調べる.

        x:要素
        """
        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):
        """要素x,yを同一視する.

        x,y:要素
        """
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.rank[x]>self.rank[y]:
            self.parents[x] += self.parents[y]
            self.parents[y] = x
        else:
            self.parents[y] += self.parents[x]
            self.parents[x] = y

            if self.rank[x]==self.rank[y]:
                self.rank[y]+=1

    def size(self, x):
        """要素xの属している要素の数.

        x:要素
        """
        return -self.parents[self.find(x)]

    def same(self, x, y):
        """要素x,yは同一視されているか?

        x,y:要素
        """
        return self.find(x) == self.find(y)

    def members(self, x):
        """要素xが属している族の要素.
        ※族の要素の個数が欲しいときはsizeを使うこと!!

        x:要素
        """
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        """族の名前のリスト
        """
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        """族の個数
        """
        return len(self.roots())

    def all_group_members(self):
        """全ての族の出力
        """
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
#================================================
N=int(input())
A=[0]*(N+1)
Inv_x=[0]*(N+1)
Inv_y=[0]*(N+1)

for i in range(N):
    x,y=map(int,input().split())
    A[x]=y
    Inv_x[x]=Inv_y[y]=i+1

b=0
B=[0]*(N+1)
for k in range(N,0,-1):
    B[k]=b=max(A[k],b)

U=Union_Find(N+1)
for k in range(1,N+1):
    if A[k]<B[k]:
        print(A[k],B[k])
        p=Inv_y[A[k]]
        q=Inv_y[B[k]]
        U.union(p,q)

X=[U.size(i) for i in range(1,N+1)]
print("\n".join(map(str,X)))