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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import(
  "fmt"
)

func main(){
  var n,tmp_x, tmp_y int
  var town_x []int
  var town_y []int
  fmt.Scanf("%d", &n)
  for i:= 0 ; i < n ; i++{
    fmt.Scanf("%d %d", &tmp_x, &tmp_y)
    town_x = append(town_x, tmp_x)
    town_y = append(town_y, tmp_y)
  }
  
  //can := make([][]bool, n)
  var group []int
  var cnt []int
  for m:=0; m<n; m++{
    group = append(group, -1)
    cnt = append(cnt, 1)
  }
  
  for j:=0; j<n; j++{
    if (group[j] != -1){
      //fmt.Printf("Town %d in Group %d/", j, group[j])
      continue
    }else{
        group[j] = j
    }
    for k:= 0; k<n; k++{
      if (group[k] != -1){
        continue
      }else if(k != j &&
         ((town_x[k] > town_x[j] && town_y[k] > town_y[j]) ||
          (town_x[k] < town_x[j] && town_y[k] < town_y[j]) )){
          group[k] = group[j] //kとjがつながってる場合にJを同じグループにする
          cnt[group[k]]++//group j の個数を増やす
      }
    }
  }
  
  //fmt.Println("Group:", group)
  //fmt.Println("Cnt:", cnt)
  for l:=0 ; l< n; l++{
  	fmt.Println(cnt[group[l]])
  }
}