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/all>
using namespace std;
using namespace atcoder;
using ll=long long;
template<class T,class U> inline bool chmin(T&x,U y){if(x>y){x=y;return true;}return false;}
template<class T,class U> inline bool chmax(T&x,U y){if(x<y){x=y;return true;}return false;}

void solve(){
    int n;
    cin>>n;
    vector<int> x(n),y(n),p(n);
    for(int i{};i<(n);++i){
        cin>>x[i]>>y[i];
    }
    iota(p.begin(),p.end(),0);
    sort(p.begin(),p.end(),[&](int i,int j){
        return x[i]<x[j];
    });
    dsu d(n);
    set<pair<int,int>> st;
    for(auto&i:p){
        auto it=st.lower_bound(make_pair(y[i],i));
        for(auto itt=st.begin();itt!=it;){
            d.merge(i,itt->second);
            itt=st.erase(itt);
        }
        st.emplace(y[i],i);
    }
    for(int i{};i<(n);++i){
        cout<<d.size(i)<<'\n';
    }
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
}