#include <bits/stdc++.h>
using namespace std;
bool cmp1(pair<int, int>& a, pair<int, int>& b) { return b.second < a.second; }
bool cmp2(pair<int, int>& a, pair<int, int>& b) { return b.second > a.second; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i].first >> v[i].second;
  vector<pair<int, int>> x, y;
  vector<int> r(2 * n + 1);
  for (int i = 0; i < n; ++i) {
    if (v[i].first < v[i].second)
      x.push_back(v[i]);
    else
      y.push_back(v[i]);
    r[v[i].first] = i + 1;
  }
  sort(x.begin(), x.end(), cmp1);
  sort(y.begin(), y.end(), cmp2);
  vector<int> id1, id2;
  int lst = 1e9;
  for (int i = 0; i < (int)x.size(); ++i) {
    if (x[i].first < lst) {
      id1.push_back(i);
      lst = x[i].second;
    }
  }
  lst = 0;
  for (int i = 0; i < (int)y.size(); ++i) {
    if (y[i].first > lst) {
      id2.push_back(i);
      lst = y[i].second;
    }
  }
  if (id1.size() > id2.size()) {
    cout << id1.size() << '\n';
    for (auto p : id1) {
      cout << r[x[p].first] << ' ';
    }
    cout << '\n';
  } else {
    cout << id2.size() << '\n';
    for (auto p : id2) cout << r[y[p].first] << ' ';
    cout << '\n';
  }
}

### 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;
bool cmp1(pair<int, int>& a, pair<int, int>& b) { return b.second < a.second; }
bool cmp2(pair<int, int>& a, pair<int, int>& b) { return b.second > a.second; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i].first >> v[i].second;
  vector<pair<int, int>> x, y;
  vector<int> r(2 * n + 1);
  for (int i = 0; i < n; ++i) {
    if (v[i].first < v[i].second)
      x.push_back(v[i]);
    else
      y.push_back(v[i]);
    r[v[i].first] = i + 1;
  }
  sort(x.begin(), x.end(), cmp1);
  sort(y.begin(), y.end(), cmp2);
  vector<int> id1, id2;
  int lst = 1e9;
  for (int i = 0; i < (int)x.size(); ++i) {
    if (x[i].first < lst) {
      id1.push_back(i);
      lst = x[i].second;
    }
  }
  lst = 0;
  for (int i = 0; i < (int)y.size(); ++i) {
    if (y[i].first > lst) {
      id2.push_back(i);
      lst = y[i].second;
    }
  }
  if (id1.size() > id2.size()) {
    cout << id1.size() << '\n';
    for (auto p : id1) {
      cout << r[x[p].first] << ' ';
    }
    cout << '\n';
  } else {
    cout << id2.size() << '\n';
    for (auto p : id2) cout << r[y[p].first] << ' ';
    cout << '\n';
  }
}

```