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:
import java.util.*;
import java.io.*;
public class Main {
public static void main(String... bla) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
ArrayList<Pair1> ps1 = new ArrayList<>();
ArrayList<Pair2> ps2 = new ArrayList<>();
for(int i = 0; i < n; i++){
StringTokenizer st = new StringTokenizer(br.readLine());
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
if(a > b) ps1.add(new Pair1(a, b, i+1));
else ps2.add(new Pair2(a, b, i+1));
}
if(ps1.size() > ps2.size()){
Collections.sort(ps1);
System.out.println(ps1.size());
StringBuilder sb = new StringBuilder();
for(Pair1 p: ps1){
sb.append(p.index+" ");
}
System.out.println(sb.toString().trim());
}
else{
Collections.sort(ps2);
System.out.println(ps2.size());
StringBuilder sb = new StringBuilder();
for(Pair2 p: ps2){
sb.append(p.index+" ");
}
System.out.println(sb.toString().trim());
}
}
static class Pair1 implements Comparable<Pair1>{
int a, b; int index;
public Pair1(int a, int b, int index){
this.a = a; this.b = b; this.index = index;
}
public int compareTo(Pair1 p){
return a - p.a;
}
}
static class Pair2 implements Comparable<Pair2>{
int a, b; int index;
public Pair2(int a, int b, int index){
this.a = a; this.b = b; this.index = index;
}
public int compareTo(Pair2 p){
return p.a - a;
}
}
}