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;
typedef long long ll;

int main() {
	int n;
    cin >> n;
    map<pair<int, int>, int> index;
    set<pair<int, int>> nodes;
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        nodes.emplace(x, y);
        index[make_pair(x, y)] = i;
    }

    dsu uf(n);
    while(nodes.size()) {
        auto p = *nodes.begin();
        nodes.erase(nodes.begin());
        int ci = index[p];
        int cx = p.first, cy = p.second;
        for(auto to : nodes) {
            int tx = to.first, ty = to.second;
            if((tx < cx && ty < cy) || (cx < tx && cy < ty)) {
                int ti = index[make_pair(tx, ty)];
                uf.merge(ci, ti);
                nodes.erase(nodes.find(to));
            }
        }
    }

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