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;
const int INF = 1e9;
const int maxn = 3e5 + 100;
bool ok(const std::vector<long long> &a, const std::vector<long long> &b,
        int ind_b, int k) {
  int need = INF;
  for (int i = 0; i < a.size() && a[i] <= b[ind_b]; ++i) {
    auto it = lower_bound(b.begin(), b.begin() + ind_b, a[i]);
    int del_b = ind_b - (it - b.begin());
    int del_a = i;
    need = min(need, del_a + del_b);
  }
  return need <= k;
}
vector<pair<int, int>> ans;
int obr[maxn];
void swap_p(vector<int> &p, int a, int b) {
  if (a == b) {
    return;
  }
  ans.emplace_back(a, b);
  swap(p[a], p[b]);
  obr[p[a]] = a;
  obr[p[b]] = b;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> pp;
  for (int i = 1; i <= n; ++i) {
    int a, b;
    cin >> a >> b;
    pp.emplace_back(a, b);
  }
  vector<int> ind_increasing;
  vector<int> ind_decreasing;
  for (int i = 0; i < n; ++i) {
    if (pp[i].first < pp[i].second) {
      ind_increasing.push_back(i);
    } else {
      ind_decreasing.push_back(i);
    }
  }
  if (ind_increasing.size() > ind_decreasing.size()) {
    vector<pair<int, int>> sorted_ind;
    for (int ind : ind_increasing) {
      sorted_ind.push_back({pp[ind].first, ind});
    }
    sort(sorted_ind.begin(), sorted_ind.end());
    cout << ind_increasing.size() << "\n";
    for (int i = sorted_ind.size() - 1; i >= 0; --i) {
      cout << sorted_ind[i].second + 1 << " ";
    }
  } else {
    vector<pair<int, int>> sorted_ind;
    for (int ind : ind_decreasing) {
      sorted_ind.push_back({pp[ind].first, ind});
    }
    sort(sorted_ind.begin(), sorted_ind.end());
    cout << ind_decreasing.size() << "\n";
    for (int i = 0; i < sorted_ind.size(); ++i) {
      cout << sorted_ind[i].second + 1 << " ";
    }
  }
}