Problem: Phoenix is trying to take a photo of his n friends with labels 1, 2, ..., n who are lined up in a row in a special order. But before he can take the photo, his friends get distracted by a duck and mess up their order.

Now, Phoenix must restore the order but he doesn't remember completely! He only remembers that the i-th friend from the left had a label between a_i and b_i inclusive. Does there exist a unique way to order his friends based of his memory? 

Input

The first line contains one integer n (1 ≤ n ≤ 2⋅10^5) — the number of friends.

The i-th of the next n lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n) — Phoenix's memory of the i-th position from the left.

It is guaranteed that Phoenix's memory is valid so there is at least one valid ordering.

Output

If Phoenix can reorder his friends in a unique order, print YES followed by n integers — the i-th integer should be the label of the i-th friend from the left.

Otherwise, print NO. Then, print any two distinct valid orderings on the following two lines. If are multiple solutions, print any.

Examples

Input


4
4 4
1 3
2 4
3 4


Output


YES
4 1 2 3 


Input


4
1 3
2 4
3 4
2 3


Output


NO
1 3 4 2 
1 2 4 3 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
 
public class Main {
  public static void main(String[] args) throws Exception {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out), false);
    new Main(in, out);
    out.close();
  }
 
  public Main(BufferedReader in, PrintWriter out) throws Exception {
    // read all required data
    int n = Integer.parseInt(in.readLine());
    Person[] people = new Person[n+1];
    List<Person>[] blocks = new List[n+1];
    for (int i = 1; i <= n; ++i) {
      blocks[i] = new ArrayList<>();
    }
    for (int i = 1; i <= n; ++i) {
      String[] vals = in.readLine().split(" ");
      int a = Integer.parseInt(vals[0]);
      int b = Integer.parseInt(vals[1]);
      people[i] = new Person(i, a, b);
      blocks[a].add(people[i]);
    }
 
    // find one valid list
    PriorityQueue<Person> pq = new PriorityQueue<>(1, new RightBoundComparator());
    for (int a = 1; a <= n; ++a) {
      for (Person p : blocks[a]) {
        pq.add(p);
      }
      while (pq.peek().b < a) {
        pq.poll();
      }
      Person p = pq.poll();
      p.v = p.v2 = a;
    }
 
    // find a second valid list
    boolean found = false;
    pq = new PriorityQueue<>(1, new RightBoundComparator());
    TreeSet<Person> index = new TreeSet<>(new ValueComparator());
    for (int a = 1; a <= n && !found; ++a) {
      for (Person p : blocks[a]) {
        pq.add(p);
        index.add(p);
      }
      while (pq.peek().b < a) {
        index.remove(pq.poll());
      }
      Person p = pq.poll();
      index.remove(p);

      p.v = p.a;
      Person p2 = index.ceiling(p);
      p.v = p.v2;
      if (p2 != null && p2.v <= p.b) {
        found = true;
        p.v2 = p2.v;
        p2.v2 = p.v;
      }
    }
 
    if (!found) {
      out.println("YES");
      for (int i = 1; i <= n; ++i) {
        out.printf("%d ", people[i].v);
      }
      out.println();
    } else {
      out.println("NO");
      for (int i = 1; i <= n; ++i) {
        out.printf("%d ", people[i].v);
      }
      out.println();
      for (int i = 1; i <= n; ++i) {
        out.printf("%d ", people[i].v2);
      }
      out.println();
    }
  }

  class RightBoundComparator implements Comparator<Person> {
    @Override
    public int compare(Person p1, Person p2) {
      if (p1.b != p2.b) return Integer.compare(p1.b, p2.b);
      return Integer.compare(p1.i, p2.i);
    }
  }

  class ValueComparator implements Comparator<Person> {
    @Override
    public int compare(Person p1, Person p2) {
      return Integer.compare(p1.v, p2.v);
    }
  }

  class Person {
    int a, b, i, v, v2;
 
    public Person(int i, int a, int b) {
      this.i = i;
      this.a = a;
      this.b = b;
    }
  }
}