#include <bits/stdc++.h>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool sort1(const pair<pair<int, int>, int> &a,
           const pair<pair<int, int>, int> &b) {
  return (a.first.second < b.first.second);
}
bool sort2(const pair<pair<int, int>, int> &a,
           const pair<pair<int, int>, int> &b) {
  return (a.first.first > b.first.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<pair<int, int>, int>> a, b;
    for (int i = 0; i < n; i++) {
      long long c, d;
      cin >> c >> d;
      if (c > d)
        a.push_back(make_pair(make_pair(c, d), i + 1));
      else
        b.push_back(make_pair(make_pair(c, d), i + 1));
    }
    sort(a.begin(), a.end(), sort1);
    sort(b.begin(), b.end(), sort2);
    if (a.size() > b.size()) {
      cout << a.size() << "\n";
      for (auto z : a) cout << z.second << " ";
    } else {
      cout << b.size() << "\n";
      for (auto z : b) cout << z.second << " ";
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool sort1(const pair<pair<int, int>, int> &a,
           const pair<pair<int, int>, int> &b) {
  return (a.first.second < b.first.second);
}
bool sort2(const pair<pair<int, int>, int> &a,
           const pair<pair<int, int>, int> &b) {
  return (a.first.first > b.first.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<pair<int, int>, int>> a, b;
    for (int i = 0; i < n; i++) {
      long long c, d;
      cin >> c >> d;
      if (c > d)
        a.push_back(make_pair(make_pair(c, d), i + 1));
      else
        b.push_back(make_pair(make_pair(c, d), i + 1));
    }
    sort(a.begin(), a.end(), sort1);
    sort(b.begin(), b.end(), sort2);
    if (a.size() > b.size()) {
      cout << a.size() << "\n";
      for (auto z : a) cout << z.second << " ";
    } else {
      cout << b.size() << "\n";
      for (auto z : b) cout << z.second << " ";
    }
  }
}

```