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: Accepted)

#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

class UnionFind {
	vector<int>par, sz;
public:
	UnionFind() {}
	UnionFind(int n) {
		par = sz = vector<int>(n);
		for (int i = 0; i < n; i++) {
			par[i] = i;
			sz[i] = 1;
		}
	}
	int find(int x) {
		if (par[x] == x)return x;
		return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y)return;
		if (sz[x] > sz[y]) {
			par[y] = x;
			sz[x] += sz[y];
		}
		else {
			par[x] = y;
			sz[y] += sz[x];
		}
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
	int size(int x) {
		return sz[find(x)];
	}
};

int x[300000],y[300000];
int p[300000];

int main(){
	int n;cin>>n;
	rep(i,n){
		scanf("%d%d",&x[i],&y[i]);x[i]--;y[i]--;
		p[y[i]]=x[i];
	}
	stack<P>st;
	UnionFind uf(n);
	for(int i=n-1;i>=0;i--){
		int Max=p[i];
		while(!st.empty()&&st.top().first>p[i]){
			Max=max(Max,st.top().first);
			uf.unite(i,st.top().second);
			st.pop();
		}
		st.push(P(Max,i));
	}
	rep(i,n){
		printf("%d\n",uf.size(y[i]));
	}
}