#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > a, b;
vector<int> temp;
bool c1(int i, int j) { return a[temp[i]].first > a[temp[j]].first; }
bool c2(int i, int j) { return b[temp[i]].first < b[temp[j]].first; }
int main() {
  int n;
  cin >> n;
  vector<int> A, B;
  temp.assign(n, -1);
  int aa, bb;
  aa = bb = 0;
  for (int i = 0 - (0 > n); i != n - (0 > n); i += 1 - 2 * (0 > n)) {
    int l, r;
    cin >> l >> r;
    if (l > r) {
      b.push_back({r, l});
      B.push_back(i);
      temp[i] = bb++;
    } else {
      a.push_back({r, l});
      A.push_back(i);
      temp[i] = aa++;
    }
  }
  sort((A).begin(), (A).end(), c1);
  sort((B).begin(), (B).end(), c2);
  if ((A).size() > (B).size()) {
    cout << (A).size() << "\n";
    for (int i = 0 - (0 > (A).size()); i != (A).size() - (0 > (A).size());
         i += 1 - 2 * (0 > (A).size()))
      cout << A[i] + 1 << " ";
    cout << "\n";
  } else {
    cout << (B).size() << "\n";
    for (int i = 0 - (0 > (B).size()); i != (B).size() - (0 > (B).size());
         i += 1 - 2 * (0 > (B).size()))
      cout << B[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
vector<pair<int, int> > a, b;
vector<int> temp;
bool c1(int i, int j) { return a[temp[i]].first > a[temp[j]].first; }
bool c2(int i, int j) { return b[temp[i]].first < b[temp[j]].first; }
int main() {
  int n;
  cin >> n;
  vector<int> A, B;
  temp.assign(n, -1);
  int aa, bb;
  aa = bb = 0;
  for (int i = 0 - (0 > n); i != n - (0 > n); i += 1 - 2 * (0 > n)) {
    int l, r;
    cin >> l >> r;
    if (l > r) {
      b.push_back({r, l});
      B.push_back(i);
      temp[i] = bb++;
    } else {
      a.push_back({r, l});
      A.push_back(i);
      temp[i] = aa++;
    }
  }
  sort((A).begin(), (A).end(), c1);
  sort((B).begin(), (B).end(), c2);
  if ((A).size() > (B).size()) {
    cout << (A).size() << "\n";
    for (int i = 0 - (0 > (A).size()); i != (A).size() - (0 > (A).size());
         i += 1 - 2 * (0 > (A).size()))
      cout << A[i] + 1 << " ";
    cout << "\n";
  } else {
    cout << (B).size() << "\n";
    for (int i = 0 - (0 > (B).size()); i != (B).size() - (0 > (B).size());
         i += 1 - 2 * (0 > (B).size()))
      cout << B[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}

```