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;
bool cmp1(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second > b.second.second;
}
bool cmp2(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second < b.second.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, a, b;
  cin >> n;
  vector<pair<int, pair<int, int> > > naik, turun;
  vector<int> ans1, ans2;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    if (a < b)
      naik.push_back(make_pair(i + 1, make_pair(a, b)));
    else
      turun.push_back(make_pair(i + 1, make_pair(a, b)));
  }
  if (!naik.empty()) {
    sort(naik.begin(), naik.end(), cmp1);
    ans1.push_back(naik[0].first);
    int tmp1 = naik[0].second.second;
    for (int i = 1; i < naik.size(); i++) {
      if (tmp1 > naik[i].second.first) {
        ans1.push_back(naik[i].first);
        tmp1 = naik[i].second.second;
      }
    }
  }
  if (!turun.empty()) {
    sort(turun.begin(), turun.end(), cmp2);
    ans2.push_back(turun[0].first);
    int tmp2 = turun[0].second.second;
    for (int i = 1; i < turun.size(); i++) {
      if (tmp2 < turun[i].second.first) {
        ans2.push_back(turun[i].first);
        tmp2 = turun[i].second.second;
      }
    }
  }
  if (ans1.size() > ans2.size()) {
    cout << ans1.size() << endl;
    for (auto i : ans1) cout << i << " ";
    cout << endl;
  } else {
    cout << ans2.size() << endl;
    for (auto i : ans2) cout << i << " ";
    cout << endl;
  }
  return 0;
}