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;
priority_queue<long long int, vector<long long int>, greater<long long int>> ti;
vector<long long int> p[100005], a(2000005), f(200005, 0), d(200005, 0),
    b(1000005);
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<pair<pair<long long int, long long int>, long long int>> p, q;
  long long int x, y;
  for (long long int i = 1; i <= n; i++) {
    cin >> x >> y;
    if (x > y) {
      p.push_back({{y, x}, i});
    }
    if (x < y) {
      q.push_back({{x, y}, i});
    }
  }
  if (p.size() > q.size()) {
    cout << p.size() << "\n";
    sort(p.begin(), p.end());
    for (long long int i = 0; i < p.size(); i++) {
      cout << p[i].second << " ";
    }
    cout << "\n";
  } else {
    cout << q.size() << "\n";
    sort(q.rbegin(), q.rend());
    for (long long int i = 0; i < q.size(); i++) {
      cout << q[i].second << " ";
    }
    cout << "\n";
  }
}