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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

#define rep(i,x,y) for(int i=x,i##end=y;i<=i##end;++i)
#define _rep(i,x,y) for(int i=x,i##end=y;i>=i##end;--i)
#define ll long long
#define pk push_back
typedef std::pair<int,int> pii;
#define N 500005
int fa[N],sz[N],x[N],y[N],cpy[N],a[N],mp[N];
std::set<int> s;
int n;
inline int ancestor(int x){
	return fa[x]==x?x:fa[x]=ancestor(fa[x]);
}
inline void merge(int x,int y){
	x=ancestor(x),y=ancestor(y);
	if(x!=y) fa[x]=y,sz[y]+=sz[x];
}
int main(){
	std::ios::sync_with_stdio(0),std::cin.tie(0),std::cout.tie(0);
	std::cin>>n;
	rep(i,1,n) std::cin>>x[i]>>y[i],mp[x[i]]=y[i];
	rep(i,1,n) sz[i]=1,fa[i]=i;
	s.insert(mp[1]);
	rep(i,2,n){
		std::set<int>::iterator it=s.lower_bound(mp[i]);
		if(it==s.begin()){
			s.insert(mp[i]);
			continue;
		}
		int cnt=0;
		for(std::set<int>::iterator j=s.begin();j!=it;j++)
			cpy[++cnt]=*j;
		rep(j,1,cnt-1) merge(cpy[j+1],cpy[j]);
		merge(mp[i],cpy[1]),s.insert(cpy[1]);
	}
	rep(i,1,n) std::cout<<sz[ancestor(y[i])]<<'\n';
	return 0;
}