#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> fen;
void update(int i, int n, pair<int, int> el) {
  while (i < n) {
    if (fen[i].first < el.first) fen[i] = el;
    i = (i | (i + 1));
  }
}
pair<int, int> get(int i) {
  pair<int, int> ans = fen[i];
  while (i >= 0) {
    if (fen[i].first > ans.first) {
      ans = fen[i];
    }
    i = (i & (i + 1)) - 1;
  }
  return ans;
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second > b.first.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    fen.push_back({0, -1});
    fen.push_back({0, -1});
  }
  vector<pair<pair<int, int>, int>> m(n);
  int ans = 0;
  int ind = 0;
  vector<int> last(n, 0);
  for (int i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    m[i].first.first = a;
    m[i].first.second = b;
    m[i].second = i;
  }
  sort(m.begin(), m.end(), cmp);
  for (int i = 0; i < n; ++i) {
    if (m[i].first.first > m[i].first.second) continue;
    pair<int, int> cur_ans = get(2 * n - 1 - m[i].first.first);
    last[m[i].second] = cur_ans.second;
    int cc1 = cur_ans.first + 1;
    int cc2 = m[i].second;
    if (ans < cur_ans.first + 1) {
      ans = cur_ans.first + 1;
      ind = m[i].second;
    }
    update(2 * n - 1 - m[i].first.second, 2 * n, {cc1, cc2});
  }
  for (int i = 0; i < 2 * n; ++i) {
    fen[i] = {0, -1};
  }
  for (int i = 0; i < n; ++i) {
    m[i].first.first = 2 * n - 1 - m[i].first.first;
    m[i].first.second = 2 * n - 1 - m[i].first.second;
  }
  sort(m.begin(), m.end(), cmp);
  for (int i = 0; i < n; ++i) {
    if (m[i].first.first > m[i].first.second) continue;
    pair<int, int> cur_ans = get(2 * n - 1 - m[i].first.first);
    last[m[i].second] = cur_ans.second;
    int cc1 = cur_ans.first + 1;
    int cc2 = m[i].second;
    if (ans < cur_ans.first + 1) {
      ans = cur_ans.first + 1;
      ind = m[i].second;
    }
    update(2 * n - 1 - m[i].first.second, 2 * n, {cc1, cc2});
  }
  cout << ans << '\n';
  vector<int> ids;
  while (ind != -1) {
    ids.push_back(ind);
    ind = last[ind];
  }
  reverse(ids.begin(), ids.end());
  for (int i = 0; i < ans; ++i) cout << ids[i] + 1 << ' ';
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
vector<pair<int, int>> fen;
void update(int i, int n, pair<int, int> el) {
  while (i < n) {
    if (fen[i].first < el.first) fen[i] = el;
    i = (i | (i + 1));
  }
}
pair<int, int> get(int i) {
  pair<int, int> ans = fen[i];
  while (i >= 0) {
    if (fen[i].first > ans.first) {
      ans = fen[i];
    }
    i = (i & (i + 1)) - 1;
  }
  return ans;
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second > b.first.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    fen.push_back({0, -1});
    fen.push_back({0, -1});
  }
  vector<pair<pair<int, int>, int>> m(n);
  int ans = 0;
  int ind = 0;
  vector<int> last(n, 0);
  for (int i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    m[i].first.first = a;
    m[i].first.second = b;
    m[i].second = i;
  }
  sort(m.begin(), m.end(), cmp);
  for (int i = 0; i < n; ++i) {
    if (m[i].first.first > m[i].first.second) continue;
    pair<int, int> cur_ans = get(2 * n - 1 - m[i].first.first);
    last[m[i].second] = cur_ans.second;
    int cc1 = cur_ans.first + 1;
    int cc2 = m[i].second;
    if (ans < cur_ans.first + 1) {
      ans = cur_ans.first + 1;
      ind = m[i].second;
    }
    update(2 * n - 1 - m[i].first.second, 2 * n, {cc1, cc2});
  }
  for (int i = 0; i < 2 * n; ++i) {
    fen[i] = {0, -1};
  }
  for (int i = 0; i < n; ++i) {
    m[i].first.first = 2 * n - 1 - m[i].first.first;
    m[i].first.second = 2 * n - 1 - m[i].first.second;
  }
  sort(m.begin(), m.end(), cmp);
  for (int i = 0; i < n; ++i) {
    if (m[i].first.first > m[i].first.second) continue;
    pair<int, int> cur_ans = get(2 * n - 1 - m[i].first.first);
    last[m[i].second] = cur_ans.second;
    int cc1 = cur_ans.first + 1;
    int cc2 = m[i].second;
    if (ans < cur_ans.first + 1) {
      ans = cur_ans.first + 1;
      ind = m[i].second;
    }
    update(2 * n - 1 - m[i].first.second, 2 * n, {cc1, cc2});
  }
  cout << ans << '\n';
  vector<int> ids;
  while (ind != -1) {
    ids.push_back(ind);
    ind = last[ind];
  }
  reverse(ids.begin(), ids.end());
  for (int i = 0; i < ans; ++i) cout << ids[i] + 1 << ' ';
  return 0;
}

```