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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            TaskD.Item[] items = new TaskD.Item[n];
            for (int i = 0; i < n; ++i) {
                items[i] = new TaskD.Item(i + 1);
                items[i].a = in.nextInt() - 1;
                items[i].b = in.nextInt() - 1;
            }

            List<TaskD.Item> res = new ArrayList<>();

            for (int iter = 0; iter < 2; ++iter) {
                TreeSet<TaskD.Item> byA = new TreeSet<>(new Comparator<TaskD.Item>() {

                    public int compare(TaskD.Item o1, TaskD.Item o2) {
                        return o1.a - o2.a;
                    }
                });
                TreeSet<TaskD.Item> byB = new TreeSet<>(new Comparator<TaskD.Item>() {

                    public int compare(TaskD.Item o1, TaskD.Item o2) {
                        return o1.b - o2.b;
                    }
                });
                for (TaskD.Item item : items) {
                    if (item.a > item.b) {
                        byA.add(item);
                        byB.add(item);
                    }
                }
                List<TaskD.Item> cur = new ArrayList<>();
                while (!byB.isEmpty()) {
                    TaskD.Item next = byB.pollFirst();
                    byA.remove(next);
                    while (!byA.isEmpty()) {
                        TaskD.Item cand = byA.first();
                        if (cand.a < next.b) {
                            byA.remove(cand);
                            byB.remove(cand);
                        } else {
                            break;
                        }
                    }
                    cur.add(next);
                }
                if (cur.size() > res.size()) {
                    res = cur;
                    if (iter == 1) {
                        Collections.reverse(res);
                    }
                }
                for (TaskD.Item item : items) {
                    int t = item.a;
                    item.a = item.b;
                    item.b = t;
                }
            }
            out.println(res.size());
            for (TaskD.Item item : res) out.print(item.index + " ");
            out.println();
        }

        static class Item {
            int index;
            int a;
            int b;

            public Item(int index) {
                this.index = index;
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}