Dirty Deeds Done Dirt Cheap

You are given $n$ pairs of integers $(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$. All of the integers in the pairs are distinct and are in the range from $1$ to $2 \cdot n$ inclusive.

Let’s call a sequence of integers $x_1, x_2, \ldots, x_{2k}$ good if either

• $x_1 < x_2 > x_3 < \ldots < x_{2k-2} > x_{2k-1} < x_{2k}$, or
• $x_1 > x_2 < x_3 > \ldots > x_{2k-2} < x_{2k-1} > x_{2k}$.

You need to choose a subset of distinct indices $i_1, i_2, \ldots, i_t$ and their order in a way that if you write down all numbers from the pairs in a single sequence (the sequence would be $a_{i_1}, b_{i_1}, a_{i_2}, b_{i_2}, \ldots, a_{i_t}, b_{i_t}$), this sequence is good.

What is the largest subset of indices you can choose? You also need to construct the corresponding index sequence $i_1, i_2, \ldots, i_t$.

Input

The first line contains single integer $n$ ($2 \leq n \leq 3 \cdot 10^5$) — the number of pairs.

Each of the next $n$ lines contain two numbers — $a_i$ and $b_i$ ($1 \le a_i, b_i \le 2 \cdot n$) — the elements of the pairs.

It is guaranteed that all integers in the pairs are distinct, that is, every integer from $1$ to $2 \cdot n$ 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 $i_1, i_2, \ldots, i_t$ — 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:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> a(n);
vector<int> b(n);
vector<int> lt;
vector<int> gt;
for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
if (a[i] < b[i]) {
      lt.push_back(i);
    } else {
      gt.push_back(i);
    }
  }
  vector<int> ans;
if (lt.size() > gt.size()) {
sort(lt.begin(), lt.end(), [&](int i, int j) {
return b[i] > b[j];
});
ans = lt;
} else {
sort(gt.begin(), gt.end(), [&](int i, int j) {
return a[i] < a[j];
    });
    ans = gt;
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < (int) ans.size(); i++) {
    if (i > 0) {
cout << " ";
    }
    cout << ans[i] + 1;
  }
  cout << '\n';
  return 0;
}