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: Accepted)

#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <list>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <bitset>
#include <random>
#include <complex>
#include <assert.h>

#include <atcoder/dsu.hpp>
#include <atcoder/fenwicktree.hpp>
#include <atcoder/segtree.hpp>
#include <atcoder/lazysegtree.hpp>
#include <atcoder/maxflow.hpp>
#include <atcoder/mincostflow.hpp>

using namespace std;
using namespace atcoder;

////////////////	型名など
typedef long long ll;
#define endl '\n'

////////////////	各種定数
const ll INF = 1LL << 60;
const double EPS = 1e-12;

////////////////	便利なテンプレート関数
template<typename T>
inline bool chmin(T& x, T a) {
	if (x >= a) { x = a; return true; }
	return false;
}

template<typename T>
inline bool chmax(T& x, T a) {
	if (x <= a) { x = a; return true; }
	return false;
}

////////////////	UnionFind
#define UnionFind dsu

int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(0);

	int N; cin >> N;
	vector<pair<pair<int, int>, int>> A(N);
	for (int i = 0; i < N; i++) {
		cin >> A[i].first.first >> A[i].first.second;
		A[i].first.first--;
		A[i].first.second--;
		A[i].second = i;
	}

	sort(A.begin(), A.end());

	set<pair<int, int>> st;// y座標とインデックス
	UnionFind uf(N);
	for (int i = 0; i < N; i++) {
		pair<int, int> p = { A[i].first.second, A[i].second };
		if (st.size() == 0 || st.begin()->first > p.first) {
			st.insert(p);
		}
		else {
			auto end = st.lower_bound(p);
			auto itr = st.begin();
			while (1) {
				itr++;
				if (itr == end) { break; }
				uf.merge(itr->second, p.second);
			}
			uf.merge(st.begin()->second, p.second);
		}
	}

	for (int i = 0; i < N; i++) {
		cout << uf.size(i) << endl;
	}

	return 0;
}