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.io.*;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class D {

    public static void main(String[] args) throws IOException {

        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt();
        ArrayList<Pair> inc = new ArrayList<>();
        ArrayList<Pair> dec = new ArrayList<>();
        for(int i = 0; i < n; ++i) {
            int a = sc.nextInt(), b = sc.nextInt();
            if (a < b)
                inc.add(new Pair(a, b, i + 1));
            else
                dec.add(new Pair(a, b, i + 1));
        }
        if (inc.size() > dec.size()) {
            out.println(inc.size());
            inc.sort((x, y) -> y.first - x.first);
            for(Pair p: inc)
                out.print(p.idx + " ");
        } else {
            out.println(dec.size());
            dec.sort((x, y) -> x.first - y.first);
            for(Pair p: dec)
                out.print(p.idx + " ");
        }

        out.close();
    }

    static class Pair  {
        int idx;
        int first, second;

        Pair(int a, int b, int c) {
            first = a;
            second = b;
            idx = c;
        }

    }

    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {return Integer.parseInt(next());}

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public String nextLine() throws IOException {return br.readLine();}

        public boolean ready() throws IOException {return br.ready();}


    }
}