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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i;
  cin >> n;
  vector<pair<int, pair<int, int>>> b, second;
  for (i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    if (x < y) {
      second.push_back(make_pair(i + 1, make_pair(x, y)));
    } else {
      b.push_back(make_pair(i + 1, make_pair(x, y)));
    }
  }
  if (second.size() > b.size()) {
    sort(second.begin(), second.end(),
         [&](pair<int, pair<int, int>> &k, pair<int, pair<int, int>> &l) {
           return k.second.second > l.second.second;
         });
    cout << second.size() << endl;
    for (i = 0; i < second.size(); i++) {
      cout << second[i].first << " ";
    }
  } else {
    sort(b.begin(), b.end(),
         [&](pair<int, pair<int, int>> &k, pair<int, pair<int, int>> &l) {
           return k.second.second < l.second.second;
         });
    cout << b.size() << endl;
    for (i = 0; i < b.size(); i++) {
      cout << b[i].first << " ";
    }
  }
}