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: Runtime Error)

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll,ll> P;
typedef string str;
typedef vector<P> vp;
typedef vector<string> vs;
typedef vector<bool> vb;

const ll mod=1e9+7;
const ll inf=1e18;

#define rep(i,m,n) for(ll i=m;i<n;i++)
#define repr(i,m,n) for(ll i=m-1;i>=n;i--)
#define fi first
#define se second
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define eb(x) emplace_back(x)
#define pb(x) pop_back(x)
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define sum(x) accumulate(all(x),0)
#define lb(x,a) lower_bound(all(x),a)
#define ub(x,a) upper_bound(all(x),a)
#define pc(x) __builtin_popcount(x)
#define gll greater<ll>()

struct UnionFind {
    vector<int> par,si; 

    UnionFind(int N) : par(N), si(N,1) { 
        rep(i,0,N) par[i] = i;
    }

    int root(int x) { 
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        x = root(x); 
        y = root(y); 
        if (x == y) return; 
        si[y]+=si[x];
        par[x] = y; 
    }

    bool same(int x, int y) { 
      return root(x)== root(y);
    }
    int size(int x){
       return si[root(x)];
    }

};


void solve(){
 ll n;
  cin >> n;
  vvll a(n,vll(n,0));
  vll x(n),y(n);
  set<ll> st;
  rep(i,0,n){
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    st.insert(i);
  }
  
 UnionFind u(n);
  while(st.size()){
   ll z=*st.begin();
   st.erase(z);
   for(auto c:st){
    if((x[z]<x[c]&&y[z]<y[c])||(x[z]>x[c]&&y[z]>y[c])) {
      u.unite(z,c);
      st.erase(c); 
    }
   }
    
  }
  
  rep(i,0,n) cout << u.size(i) << endl;
  
}
int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);
  solve();
}