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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long spf[15000001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 15000001; i++) spf[i] = i;
  for (long long i = 4; i < 15000001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 15000001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 15000001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
vector<long long> getFactorization(long long x) {
  vector<long long> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  sieve();
  long long n;
  cin >> n;
  vector<pair<pair<long long, long long>, long long> > inc, dec;
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    if (a > b)
      inc.push_back(make_pair(make_pair(b, a), i));
    else
      dec.push_back(make_pair(make_pair(a, b), i));
  }
  sort(inc.begin(), inc.end());
  sort(dec.begin(), dec.end());
  if (inc.size() > dec.size()) {
    cout << inc.size() << '\n';
    for (long long i = 0; i < inc.size(); i++) cout << inc[i].second + 1 << " ";
    cout << '\n';
  } else {
    cout << dec.size() << '\n';
    long long k = dec.size();
    for (long long j = k - 1; j >= 0; j--) cout << dec[j].second + 1 << " ";
    cout << '\n';
  }
}