#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n;
  cin >> n;
  vector<pair<ll, ll> > a(n);
  for (auto &p : a) cin >> p.first >> p.second;
  vector<pair<ll, ll> > p;
  for (ll i = 0; i < n; i++)
    if (a[i].first > a[i].second) p.emplace_back(a[i].second, i);
  sort(begin(p), end(p));
  vector<ll> res1;
  if (ll((p).size()) > 1) res1.push_back(p[0].second);
  for (ll i = 1; i < ll((p).size()); i++) {
    if (a[res1[i - 1]].second < a[p[i].second].first)
      res1.push_back(p[i].second);
  }
  p.clear();
  for (ll i = 0; i < n; i++)
    if (a[i].first < a[i].second) p.emplace_back(a[i].second, i);
  sort(begin(p), end(p), greater<pair<ll, ll> >());
  vector<ll> res2;
  if (ll((p).size()) > 0) res2.push_back(p[0].second);
  for (ll i = 1; i < ll((p).size()); i++) {
    if (a[res2[i - 1]].second > a[p[i].second].first)
      res2.push_back(p[i].second);
  }
  if (ll((res1).size()) > ll((res2).size())) {
    cout << ll((res1).size()) << endl;
    for (ll i = 0; i < ll((res1).size()); i++) {
      cout << res1[i] + 1 << " \n"[i == ll((res1).size()) - 1];
    }
  } else {
    cout << ll((res2).size()) << endl;
    for (ll i = 0; i < ll((res2).size()); i++) {
      cout << res2[i] + 1 << " \n"[i == ll((res2).size()) - 1];
    }
  }
}

### Prompt

Generate a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n;
  cin >> n;
  vector<pair<ll, ll> > a(n);
  for (auto &p : a) cin >> p.first >> p.second;
  vector<pair<ll, ll> > p;
  for (ll i = 0; i < n; i++)
    if (a[i].first > a[i].second) p.emplace_back(a[i].second, i);
  sort(begin(p), end(p));
  vector<ll> res1;
  if (ll((p).size()) > 1) res1.push_back(p[0].second);
  for (ll i = 1; i < ll((p).size()); i++) {
    if (a[res1[i - 1]].second < a[p[i].second].first)
      res1.push_back(p[i].second);
  }
  p.clear();
  for (ll i = 0; i < n; i++)
    if (a[i].first < a[i].second) p.emplace_back(a[i].second, i);
  sort(begin(p), end(p), greater<pair<ll, ll> >());
  vector<ll> res2;
  if (ll((p).size()) > 0) res2.push_back(p[0].second);
  for (ll i = 1; i < ll((p).size()); i++) {
    if (a[res2[i - 1]].second > a[p[i].second].first)
      res2.push_back(p[i].second);
  }
  if (ll((res1).size()) > ll((res2).size())) {
    cout << ll((res1).size()) << endl;
    for (ll i = 0; i < ll((res1).size()); i++) {
      cout << res1[i] + 1 << " \n"[i == ll((res1).size()) - 1];
    }
  } else {
    cout << ll((res2).size()) << endl;
    for (ll i = 0; i < ll((res2).size()); i++) {
      cout << res2[i] + 1 << " \n"[i == ll((res2).size()) - 1];
    }
  }
}

```