#include <bits/stdc++.h>
using namespace std;
const int N = 600600;
int n;
int a[N], b[N];
pair<int, int> ft[N];
pair<int, int> f[N];
void reset() {
  for (int i = 1; i <= n + n; ++i) ft[i] = {0, 0};
}
void upd(int p, pair<int, int> v, bool rev) {
  if (rev) p = 2 * n + 1 - p;
  for (; p <= 2 * n; p += p & -p) {
    ft[p] = max(ft[p], v);
  }
}
pair<int, int> get(int p, bool rev) {
  if (rev) p = 2 * n + 1 - p;
  pair<int, int> ans = {0, 0};
  for (; p > 0; p -= p & -p) {
    ans = max(ans, ft[p]);
  }
  return ans;
}
vector<int> up() {
  reset();
  vector<int> values;
  for (int i = 1; i <= n; ++i) {
    if (a[i] < b[i]) values.push_back(i);
  }
  sort(values.begin(), values.end(), [&](int u, int v) { return b[u] > b[v]; });
  for (auto id : values) {
    f[id] = get(a[id], true);
    f[id].first++;
    upd(b[id], make_pair(f[id].first, id), true);
  }
  int id = get(1, true).second;
  vector<int> ans;
  while (id) {
    ans.push_back(id);
    id = f[id].second;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
vector<int> down() {
  reset();
  vector<int> values;
  for (int i = 1; i <= n; ++i) {
    if (a[i] > b[i]) values.push_back(i);
  }
  sort(values.begin(), values.end(), [&](int u, int v) { return b[u] < b[v]; });
  for (auto id : values) {
    f[id] = get(a[id], false);
    f[id].first++;
    upd(b[id], make_pair(f[id].first, id), false);
  }
  int id = get(2 * n, false).second;
  vector<int> ans;
  while (id) {
    ans.push_back(id);
    id = f[id].second;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i];
  auto f = ::up();
  auto g = ::down();
  if (f.size() < g.size()) f = g;
  cout << f.size() << '\n';
  for (int v : f) cout << v << ' ';
}

### 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;
const int N = 600600;
int n;
int a[N], b[N];
pair<int, int> ft[N];
pair<int, int> f[N];
void reset() {
  for (int i = 1; i <= n + n; ++i) ft[i] = {0, 0};
}
void upd(int p, pair<int, int> v, bool rev) {
  if (rev) p = 2 * n + 1 - p;
  for (; p <= 2 * n; p += p & -p) {
    ft[p] = max(ft[p], v);
  }
}
pair<int, int> get(int p, bool rev) {
  if (rev) p = 2 * n + 1 - p;
  pair<int, int> ans = {0, 0};
  for (; p > 0; p -= p & -p) {
    ans = max(ans, ft[p]);
  }
  return ans;
}
vector<int> up() {
  reset();
  vector<int> values;
  for (int i = 1; i <= n; ++i) {
    if (a[i] < b[i]) values.push_back(i);
  }
  sort(values.begin(), values.end(), [&](int u, int v) { return b[u] > b[v]; });
  for (auto id : values) {
    f[id] = get(a[id], true);
    f[id].first++;
    upd(b[id], make_pair(f[id].first, id), true);
  }
  int id = get(1, true).second;
  vector<int> ans;
  while (id) {
    ans.push_back(id);
    id = f[id].second;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
vector<int> down() {
  reset();
  vector<int> values;
  for (int i = 1; i <= n; ++i) {
    if (a[i] > b[i]) values.push_back(i);
  }
  sort(values.begin(), values.end(), [&](int u, int v) { return b[u] < b[v]; });
  for (auto id : values) {
    f[id] = get(a[id], false);
    f[id].first++;
    upd(b[id], make_pair(f[id].first, id), false);
  }
  int id = get(2 * n, false).second;
  vector<int> ans;
  while (id) {
    ans.push_back(id);
    id = f[id].second;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i];
  auto f = ::up();
  auto g = ::down();
  if (f.size() < g.size()) f = g;
  cout << f.size() << '\n';
  for (int v : f) cout << v << ' ';
}

```