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: Wrong Answer)

//雪花飄飄北風嘯嘯
//天地一片蒼茫

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ii,ll>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << " is " << x << endl

#define rep(x,start,end) for(auto x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

#define indexed_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
//change less to less_equal for non distinct pbds, but erase will bug

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

struct UFDS{
	int p[200005];
	int s[200005];
	
	UFDS(){
		rep(x,0,200005){
			p[x]=x;
			s[x]=1;
		}
	}
	
	int parent(int i){
		if (p[i]==i) return i;
		else return p[i]=parent(p[i]);
	}
	
	void unions(int i,int j){
		i=parent(i),j=parent(j);
		
		if (i==j) return;
		
		p[j]=i;
		s[i]+=s[j];
	}
} ufds;

int n;
ii pos[200005];

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);
	
	cin>>n;
	rep(x,0,n) cin>>pos[x].fi>>pos[x].se;
	
	vector<int> proc;
	rep(x,0,n) proc.push_back(x);
	
	sort(all(proc),[](int i,int j){
		return pos[i].fi<pos[j].fi;
	});
	
	vector<int> stk;
	
	rep(x,0,n){
		while (!stk.empty() && pos[stk.back()].se<pos[proc[x]].se){
			ufds.unions(stk.back(),proc[x]);
			//cout<<stk.back()<<" "<<x<<endl;
			stk.pop_back();
		}
		stk.push_back(proc[x]);
	}
	
	rep(x,0,n){
		cout<<ufds.s[ufds.parent(x)]<<endl;
	}
}