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 power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int aa[3 * 100005][2];
vector<pair<pair<long long int, long long int>, long long int> > v1, v2;
long long int cmp1(pair<pair<long long int, long long int>, long long int> a,
                   pair<pair<long long int, long long int>, long long int> b) {
  if (a.first.first < b.first.second) {
    return 0;
  }
  return 1;
}
long long int cmp2(pair<pair<long long int, long long int>, long long int> a,
                   pair<pair<long long int, long long int>, long long int> b) {
  if (a.first.first > b.first.second) {
    return 0;
  }
  return 1;
}
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> aa[i][0] >> aa[i][1];
    if (aa[i][0] < aa[i][1])
      v1.push_back(make_pair(make_pair(aa[i][0], aa[i][1]), i + 1));
    else
      v2.push_back(make_pair(make_pair(aa[i][0], aa[i][1]), i + 1));
  }
  sort((v1).begin(), (v1).end(), cmp1);
  sort((v2).begin(), (v2).end(), cmp2);
  if (v1.size() > v2.size()) {
    cout << v1.size() << endl;
    for (long long int i = 0; i < v1.size(); i++) {
      cout << v1[i].second << " ";
    }
  } else {
    cout << v2.size() << endl;
    for (long long int i = 0; i < v2.size(); i++) {
      cout << v2[i].second << " ";
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  for (long long int tc = 1; tc <= t; tc++) {
    solve();
  }
}