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.*;

public class GlobalRound3 {

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        TaskD solver = new TaskD();
        solver.solve(in, out);
        out.close();
    }
}

class Pair<A, B> {
    private A a;
    private B b;

    public Pair(A a, B b) {
        this.a = a;
        this.b = b;
    }

    //    public String getType() { return a.getClass().getSimpleName(); }

    public A getA() {
        return a;
    }

    public B getB() {
        return b;
    }

    public void setA(A a) {
        this.a = a;
    }

    public void setB(B b) {
        this.b = b;
    }

    @Override
    public int hashCode() {
        int h1 = a != null ? a.hashCode() : 0;
        int h2 = b != null ? b.hashCode() : 0;
        return 31 * h1 + h2;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return a.equals(p.a) && b.equals(p.b);
    }
}

class InputReader {
    private BufferedReader br;
    private StringTokenizer st;

    public InputReader(InputStream stream) {
        br = new BufferedReader(new InputStreamReader(stream));
    }

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

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

    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
}

class TaskD {

    public void solve(InputReader in, PrintWriter out) {
        int n = in.nextInt();

        List<Pair<Integer, Integer>> list1 = new ArrayList<>();
        List<Pair<Integer, Integer>> list2 = new ArrayList<>();
        Map<Pair<Integer, Integer>, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int a = in.nextInt();
            int b = in.nextInt();
            if (a < b)
                list1.add(new Pair<>(a, b));
            else
                list2.add(new Pair<>(a, b));
            map.put(new Pair<>(a, b), i + 1);
        }

        Collections.sort(list1, (o1, o2) -> o2.getB() - o1.getB());
        Collections.sort(list2, (o1, o2) -> o1.getA() - o2.getA());

        if (list1.size() >= list2.size()) {
            out.println(list1.size());
            for (Pair p : list1)
                out.print(map.get(p) + " ");

        } else {
            out.println(list2.size());
            for (Pair p : list2)
                out.print(map.get(p) + " ");

        }
        out.println();
    }
}