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;
const long long inf = (1ll << 62) - 1;
const long long mod = 1e9 + 7;
const long long N = 2e5 + 10;
long long bPow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % mod;
    }
    b >>= 1;
    a = (a * a) % mod;
  }
  return res % mod;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<long long> fact(N, 0ll);
void factorial() {
  fact[0] = 1, fact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
}
long long ncr(long long n, long long r) {
  if (r > n) return 0;
  long long ans = fact[n] % mod;
  ans *= bPow(fact[r], mod - 2) % mod;
  ans %= mod;
  ans *= bPow(fact[n - r], mod - 2) % mod;
  ans %= mod;
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<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.emplace_back(b, i);
    else
      dec.emplace_back(a, i);
  }
  sort(inc.rbegin(), inc.rend());
  sort(dec.begin(), dec.end());
  if (inc.size() > dec.size()) {
    cout << inc.size() << endl;
    for (auto it : inc) {
      cout << it.second + 1 << " ";
    }
  } else {
    cout << dec.size() << endl;
    for (auto it : dec) cout << it.second + 1 << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  while (tc--) {
    solve();
  }
}