#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
const int N = 3e5 + 5;
int n, A[N], B[N], F[N], Pre[N];
vector<int> I;
vector<int> solve() {
  if (I.empty()) return {};
  memset(F, 0, sizeof F);
  sort(I.begin(), I.end(), [&](int i, int j) { return B[i] > B[j]; });
  Pre[I[0]] = -1;
  F[0] = 1;
  int ansi = I[0], ans = 1;
  vector<int> Ret;
  for (int ii = 1; ii < int(I.size()); ii++) {
    int i = I[ii];
    int l = 0, r = ii;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      if (B[I[m]] > A[i])
        l = m;
      else
        r = m;
    }
    F[ii] = F[l] + 1;
    Pre[i] = I[l];
    if (F[ii] > ans) {
      ans = F[ii];
      ansi = i;
    }
  }
  while (ansi != -1) {
    Ret.push_back(ansi);
    ansi = Pre[ansi];
  }
  reverse(Ret.begin(), Ret.end());
  return Ret;
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) scanf("%d%d", &A[i], &B[i]);
  for (int i = (1); i <= (n); ++i)
    if (A[i] < B[i]) I.push_back(i);
  auto Ans1 = solve();
  I.clear();
  for (int i = (1); i <= (n); ++i)
    if (A[i] > B[i]) {
      A[i] = -A[i], B[i] = -B[i];
      I.push_back(i);
    }
  auto Ans2 = solve();
  auto Ans = Ans1;
  if (Ans2.size() > Ans1.size()) {
    Ans = Ans2;
  }
  cout << Ans.size() << endl;
  for (auto x : Ans) printf("%d ", x);
  puts("");
}

### Prompt

Please formulate 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;
using LL = long long;
using PII = pair<int, int>;
const int N = 3e5 + 5;
int n, A[N], B[N], F[N], Pre[N];
vector<int> I;
vector<int> solve() {
  if (I.empty()) return {};
  memset(F, 0, sizeof F);
  sort(I.begin(), I.end(), [&](int i, int j) { return B[i] > B[j]; });
  Pre[I[0]] = -1;
  F[0] = 1;
  int ansi = I[0], ans = 1;
  vector<int> Ret;
  for (int ii = 1; ii < int(I.size()); ii++) {
    int i = I[ii];
    int l = 0, r = ii;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      if (B[I[m]] > A[i])
        l = m;
      else
        r = m;
    }
    F[ii] = F[l] + 1;
    Pre[i] = I[l];
    if (F[ii] > ans) {
      ans = F[ii];
      ansi = i;
    }
  }
  while (ansi != -1) {
    Ret.push_back(ansi);
    ansi = Pre[ansi];
  }
  reverse(Ret.begin(), Ret.end());
  return Ret;
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) scanf("%d%d", &A[i], &B[i]);
  for (int i = (1); i <= (n); ++i)
    if (A[i] < B[i]) I.push_back(i);
  auto Ans1 = solve();
  I.clear();
  for (int i = (1); i <= (n); ++i)
    if (A[i] > B[i]) {
      A[i] = -A[i], B[i] = -B[i];
      I.push_back(i);
    }
  auto Ans2 = solve();
  auto Ans = Ans1;
  if (Ans2.size() > Ans1.size()) {
    Ans = Ans2;
  }
  cout << Ans.size() << endl;
  for (auto x : Ans) printf("%d ", x);
  puts("");
}

```