#include <bits/stdc++.h>
using namespace std;
namespace std {
template <>
struct hash<pair<int, int>> {
  size_t operator()(const pair<int, int> &x) const {
    long long P = 38923, Q = 109797901;
    return (size_t)((x.first * P + x.second) % Q);
  }
};
};  // namespace std
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void print(T t) {
  cout << t << endl;
}
template <typename T, typename... Args>
void print(T t, Args... args) {
  cout << t << " ";
  print(args...);
}
int n, a, b;
vector<pair<pair<int, int>, int>> g, l, sol;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = min((int)0, (int)n); i < max((int)0, (int)n); i += 1) {
    cin >> a >> b;
    if (a < b) {
      l.push_back({{a, b}, i});
    } else if (a > b) {
      g.push_back({{a, b}, i});
    } else {
      l.push_back({{a, b}, i});
      g.push_back({{a, b}, i});
    }
  }
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  sort(g.begin(), g.end());
  if ((int)(l).size() > (int)(g).size())
    sol = l;
  else
    sol = g;
  cout << (int)(sol).size() << "\n";
  for (int i = min((int)0, (int)(int)(sol).size());
       i < max((int)0, (int)(int)(sol).size()); i += 1)
    cout << (sol[i].second + 1) << " \n"[i == (int)(sol).size() - 1];
  return 0;
}

### Prompt

Create a solution in cpp for 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;
namespace std {
template <>
struct hash<pair<int, int>> {
  size_t operator()(const pair<int, int> &x) const {
    long long P = 38923, Q = 109797901;
    return (size_t)((x.first * P + x.second) % Q);
  }
};
};  // namespace std
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void print(T t) {
  cout << t << endl;
}
template <typename T, typename... Args>
void print(T t, Args... args) {
  cout << t << " ";
  print(args...);
}
int n, a, b;
vector<pair<pair<int, int>, int>> g, l, sol;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = min((int)0, (int)n); i < max((int)0, (int)n); i += 1) {
    cin >> a >> b;
    if (a < b) {
      l.push_back({{a, b}, i});
    } else if (a > b) {
      g.push_back({{a, b}, i});
    } else {
      l.push_back({{a, b}, i});
      g.push_back({{a, b}, i});
    }
  }
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  sort(g.begin(), g.end());
  if ((int)(l).size() > (int)(g).size())
    sol = l;
  else
    sol = g;
  cout << (int)(sol).size() << "\n";
  for (int i = min((int)0, (int)(int)(sol).size());
       i < max((int)0, (int)(int)(sol).size()); i += 1)
    cout << (sol[i].second + 1) << " \n"[i == (int)(sol).size() - 1];
  return 0;
}

```