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:
//package com.krakn.CF.D.D1148;

import java.io.*;
import java.util.*;

public class Main {

    static class Pair {
        int i, j, idx;

        Pair(int i, int j, int idx) {
            this.i = i;
            this.j = j;
            this.idx = idx;
        }

        public String toString() {
            return i + " - " + j;
        }
    }

    public static void main(String[] args) {
        FastReader fr = new FastReader();

        int n = fr.nextInt();

        int a, b;
        List<Pair> lPairs = new ArrayList<>();
        List<Pair> gPairs = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            a = fr.nextInt();
            b = fr.nextInt();
            if (a < b) lPairs.add(new Pair(a, b, i + 1));
            if (a > b) gPairs.add(new Pair(a, b, i + 1));
        }

        if (lPairs.size() >= gPairs.size()) {
            lPairs.sort((o1, o2) -> o2.i - o1.i);
            System.out.println(lPairs.size());
            for (Pair p : lPairs) System.out.print(p.idx + " ");
        } else {
            gPairs.sort((o1, o2) -> o1.i - o2.i);
            System.out.println(gPairs.size());
            for (Pair p : gPairs) System.out.print(p.idx + " ");
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

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