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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
int a[300005], b[300005];
bool comp(pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) {
  return a.first.second > b.first.second;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n;
  vector<pair<pair<int, int>, int>> A, B;
  for (i = 1; i <= n; i++) {
    cin >> x >> y;
    if (x > y)
      A.push_back({{x, y}, i});
    else
      B.push_back({{x, y}, i});
  }
  if (A.size() >= B.size()) {
    cout << A.size() << "\n";
    sort(A.begin(), A.end());
    for (i = 0; i < A.size(); i++) cout << A[i].second << " ";
    return 0;
  }
  cout << B.size() << "\n";
  sort(B.begin(), B.end(), comp);
  for (i = 0; i < B.size(); i++) cout << B[i].second << " ";
  return 0;
}