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 with AC Library v1.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <atcoder/dsu>
#include <cstdio>
using namespace atcoder;
using namespace std;
using ll=long long;
using ld=long double;

using tlll = tuple<ll,ll,ll>;


int main() {

    ll N;
    cin >> N;
    dsu Ue(N);
    vector<tuple<ll,ll,ll>> XY(N);
    for(ll i=0;i<N;i++){
        cin >> get<0>(XY[i]) >> get<1>(XY[i]);
        get<2>(XY[i])=i;
    }
    sort(XY.begin(), XY.end());
    
    for(ll i=0;i<N-1;i++)
        if(get<1>(XY[i])<get<1>(XY[i+1]))Ue.merge(get<2>(XY[i]), get<2>(XY[i+1]));
        
   for(ll i=N-1;i>0;i--)
        if(get<1>(XY[i])<get<1>(XY[i-1]))Ue.merge(get<2>(XY[i]), get<2>(XY[i-1]));
      
    sort(XY.begin(), XY.end(),[](tlll a,tlll b){return get<1>(a)<get<1>(b);});
    for(ll i=0;i<N-1;i++)
        if(get<0>(XY[i])<get<0>(XY[i+1]))Ue.merge(get<2>(XY[i]), get<2>(XY[i+1]));
    for(ll i=N-1;i>0;i--)
        if(get<0>(XY[i])<get<0>(XY[i-1]))Ue.merge(get<2>(XY[i]), get<2>(XY[i-1]));
  
  
  for(ll i=0;i<N;i++){
   cout << Ue.size(i) << endl; 
  }

    
}