#include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<pair<int, int>, int> > v1, v2;
bool cstm1(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second == y.first.second) {
    x.first.first < y.first.first;
  }
  return x.first.second > y.first.second;
}
bool cstm2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second == y.first.second) {
    x.first.first > y.first.first;
  }
  return x.first.second < y.first.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (int)1; i <= (int)n; i++) {
    int l, r;
    cin >> l >> r;
    if (l < r) {
      v1.push_back({{l, r}, i});
    } else if (l > r) {
      v2.push_back({{l, r}, i});
    }
  }
  vector<int> ans1, ans2;
  int k1 = (int)v1.size(), k2 = (int)v2.size();
  sort(v1.begin(), v1.end(), cstm1);
  sort(v2.begin(), v2.end(), cstm2);
  pair<pair<int, int>, int> last;
  for (int i = (int)0; i <= (int)k1 - 1; i++) {
    if (i == 0) {
      ans1.push_back(v1[i].second);
      last = v1[i];
    } else if (v1[i].first.first < last.first.second) {
      last = v1[i];
      ans1.push_back(v1[i].second);
    }
  }
  for (int i = (int)0; i <= (int)k2 - 1; i++) {
    if (i == 0) {
      ans2.push_back(v2[i].second);
      last = v2[i];
    } else if (v2[i].first.first > last.first.second) {
      last = v2[i];
      ans2.push_back(v2[i].second);
    }
  }
  cout << max((int)ans1.size(), (int)ans2.size()) << "\n";
  if ((int)ans1.size() >= (int)ans2.size()) {
    for (auto e : ans1) cout << e << " ";
  } else {
    for (auto e : ans2) cout << e << " ";
  }
  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;
int n;
vector<pair<pair<int, int>, int> > v1, v2;
bool cstm1(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second == y.first.second) {
    x.first.first < y.first.first;
  }
  return x.first.second > y.first.second;
}
bool cstm2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second == y.first.second) {
    x.first.first > y.first.first;
  }
  return x.first.second < y.first.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (int)1; i <= (int)n; i++) {
    int l, r;
    cin >> l >> r;
    if (l < r) {
      v1.push_back({{l, r}, i});
    } else if (l > r) {
      v2.push_back({{l, r}, i});
    }
  }
  vector<int> ans1, ans2;
  int k1 = (int)v1.size(), k2 = (int)v2.size();
  sort(v1.begin(), v1.end(), cstm1);
  sort(v2.begin(), v2.end(), cstm2);
  pair<pair<int, int>, int> last;
  for (int i = (int)0; i <= (int)k1 - 1; i++) {
    if (i == 0) {
      ans1.push_back(v1[i].second);
      last = v1[i];
    } else if (v1[i].first.first < last.first.second) {
      last = v1[i];
      ans1.push_back(v1[i].second);
    }
  }
  for (int i = (int)0; i <= (int)k2 - 1; i++) {
    if (i == 0) {
      ans2.push_back(v2[i].second);
      last = v2[i];
    } else if (v2[i].first.first > last.first.second) {
      last = v2[i];
      ans2.push_back(v2[i].second);
    }
  }
  cout << max((int)ans1.size(), (int)ans2.size()) << "\n";
  if ((int)ans1.size() >= (int)ans2.size()) {
    for (auto e : ans1) cout << e << " ";
  } else {
    for (auto e : ans2) cout << e << " ";
  }
  return 0;
}

```