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 int MAX = 101;
const long long int mod = 998244353;
const long long int INF = INT_MAX;
inline long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
inline long long int pow(long long int a, long long int n, long long int mod) {
  if (n == 0) return 1;
  long long int p = pow(a, n / 2, mod);
  p = (p % mod * p % mod) % mod;
  if (n % 2)
    return (p % mod * a % mod) % mod;
  else
    return p;
}
bool comp1(pair<long long int, long long int> &a,
           pair<long long int, long long int> &b) {
  return a.second < b.second;
}
bool comp2(pair<long long int, long long int> &a,
           pair<long long int, long long int> &b) {
  return a.second > b.second;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    vector<pair<long long int, long long int> > v(n);
    vector<pair<long long int, long long int> > v1;
    vector<pair<long long int, long long int> > v2;
    map<long long int, long long int> mp;
    for (long long int i = 0; i < n; i++) {
      cin >> v[i].first >> v[i].second;
      mp[v[i].first] = i + 1;
      if (v[i].first < v[i].second)
        v1.push_back({v[i].first, v[i].second});
      else
        v2.push_back({v[i].first, v[i].second});
    }
    sort(v1.begin(), v1.end(), comp2);
    sort(v2.begin(), v2.end(), comp1);
    if (v1.size() >= v2.size()) {
      cout << v1.size() << "\n";
      for (long long int i = 0; i < v1.size(); i++)
        cout << mp[v1[i].first] << " ";
    } else {
      cout << v2.size() << "\n";
      for (long long int i = 0; i < v2.size(); i++)
        cout << mp[v2[i].first] << " ";
    }
  }
}