Problem: You are given n pairs of integers (a_1, b_1), (a_2, b_2), …, (a_n, b_n). All of the integers in the pairs are distinct and are in the range from 1 to 2 ⋅ n inclusive.

Let's call a sequence of integers x_1, x_2, …, x_{2k} good if either 

  * x_1 < x_2 > x_3 < … < x_{2k-2} > x_{2k-1} < x_{2k}, or 
  * x_1 > x_2 < x_3 > … > x_{2k-2} < x_{2k-1} > x_{2k}. 



You need to choose a subset of distinct indices i_1, i_2, …, i_t and their order in a way that if you write down all numbers from the pairs in a single sequence (the sequence would be a_{i_1}, b_{i_1}, a_{i_2}, b_{i_2}, …, a_{i_t}, b_{i_t}), this sequence is good.

What is the largest subset of indices you can choose? You also need to construct the corresponding index sequence i_1, i_2, …, i_t.

Input

The first line contains single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of pairs.

Each of the next n lines contain two numbers — a_i and b_i (1 ≤ a_i, b_i ≤ 2 ⋅ n) — the elements of the pairs.

It is guaranteed that all integers in the pairs are distinct, that is, every integer from 1 to 2 ⋅ n is mentioned exactly once.

Output

In the first line print a single integer t — the number of pairs in the answer.

Then print t distinct integers i_1, i_2, …, i_t — the indexes of pairs in the corresponding order.

Examples

Input


5
1 7
6 4
2 10
9 8
3 5


Output


3
1 5 3


Input


3
5 4
3 2
6 1


Output


3
3 2 1

Note

The final sequence in the first example is 1 < 7 > 3 < 5 > 2 < 10.

The final sequence in the second example is 6 > 1 < 3 > 2 < 5 > 4.

Solution (in JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1148D
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         ArrayList<Pair> inc = new ArrayList<Pair>();
         ArrayList<Pair2> dec = new ArrayList<Pair2>();
         for(int i=0; i < N; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            if(a < b)
               inc.add(new Pair(a,b,i+1));
            else
               dec.add(new Pair2(a,b,i+1));
         }
         if(inc.size() >= dec.size())
         {
            Collections.sort(inc);
            StringBuilder sb = new StringBuilder();
            sb.append(inc.size()+"\n");
            for(Pair p: inc)
               sb.append(p.dex+" ");
            System.out.println(sb);
         }
         else
         {
            Collections.sort(dec);
            StringBuilder sb = new StringBuilder();
            sb.append(dec.size()+"\n");
            for(Pair2 p: dec)
               sb.append(p.dex+" ");
            System.out.println(sb);
         }
      }
   }
   class Pair implements Comparable<Pair>
   {
      public int a;
      public int b;
      public int dex;
      
      public Pair(int c, int d, int e)
      {
         a = c;
         b = d;
         dex = e;
      }
      public int compareTo(Pair oth)
      {
         return oth.b-b;
      }
   }
   class Pair2 implements Comparable<Pair2>
   {
      public int a;
      public int b;
      public int dex;
      
      public Pair2(int c, int d, int e)
      {
         a = c;
         b = d;
         dex = e;
      }
      public int compareTo(Pair2 oth)
      {
         return a-oth.a;
      }
   }