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;
long long n, x;
pair<long long, long long> a[300005];
vector<long long> v[2];
bool comp1(long long x, long long y) { return a[x].first > a[y].first; }
bool comp2(long long x, long long y) { return a[x].second < a[y].second; }
void solve() {
  cin >> n;
  for (__typeof((n + 1)) i = (1); i < (n + 1); i++) {
    cin >> a[i].first >> a[i].second;
    if (a[i].first < a[i].second)
      v[0].push_back(i);
    else
      v[1].push_back(i);
  }
  sort((v[0]).begin(), (v[0]).end(), comp1);
  sort((v[1]).begin(), (v[1]).end(), comp2);
  if ((long long)v[1].size() > (long long)v[0].size()) x = 1;
  cout << (long long)v[x].size() << "\n";
  for (auto i : v[x]) cout << i << " ";
}
void prep() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  prep();
  cout << fixed << setprecision(12);
  while (t--) solve();
  return 0;
}