#include <bits/stdc++.h>
using namespace std;
const int N = 300100;
int n, a[N], b[N];
vector<int> c;
vector<int> d;
vector<pair<pair<int, int>, int> > u;
vector<pair<pair<int, int>, int> > v;
void print(vector<int> &x) {
  printf("%d\n", (int)x.size());
  for (int i = 0; i < x.size(); i++) {
    printf("%d ", x[i] + 1);
  }
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    if (a[i] < b[i]) {
      u.push_back(make_pair(make_pair(b[i], a[i]), i));
    } else if (a[i] > b[i]) {
      v.push_back(make_pair(make_pair(b[i], a[i]), i));
    }
  }
  sort(u.rbegin(), u.rend());
  sort(v.begin(), v.end());
  if (u.size() > 0) {
    int last = u[0].first.first;
    c.push_back(u[0].second);
    for (int i = 1; i < u.size(); i++) {
      if (u[i].first.second < last) {
        last = u[i].first.first;
        c.push_back(u[i].second);
      }
    }
  }
  if (v.size() > 0) {
    int last = v[0].first.first;
    d.push_back(v[0].second);
    for (int i = 1; i < v.size(); i++) {
      if (v[i].first.second > last) {
        last = v[i].first.first;
        d.push_back(v[i].second);
      }
    }
  }
  if (c.size() > d.size()) {
    print(c);
  } else {
    print(d);
  }
  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;
const int N = 300100;
int n, a[N], b[N];
vector<int> c;
vector<int> d;
vector<pair<pair<int, int>, int> > u;
vector<pair<pair<int, int>, int> > v;
void print(vector<int> &x) {
  printf("%d\n", (int)x.size());
  for (int i = 0; i < x.size(); i++) {
    printf("%d ", x[i] + 1);
  }
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    if (a[i] < b[i]) {
      u.push_back(make_pair(make_pair(b[i], a[i]), i));
    } else if (a[i] > b[i]) {
      v.push_back(make_pair(make_pair(b[i], a[i]), i));
    }
  }
  sort(u.rbegin(), u.rend());
  sort(v.begin(), v.end());
  if (u.size() > 0) {
    int last = u[0].first.first;
    c.push_back(u[0].second);
    for (int i = 1; i < u.size(); i++) {
      if (u[i].first.second < last) {
        last = u[i].first.first;
        c.push_back(u[i].second);
      }
    }
  }
  if (v.size() > 0) {
    int last = v[0].first.first;
    d.push_back(v[0].second);
    for (int i = 1; i < v.size(); i++) {
      if (v[i].first.second > last) {
        last = v[i].first.first;
        d.push_back(v[i].second);
      }
    }
  }
  if (c.size() > d.size()) {
    print(c);
  } else {
    print(d);
  }
  return 0;
}

```