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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const long long int inf = 9e18;
const long double pi = 2 * acos(0.0);
using namespace std;
long long int power(long long int a, long long int n) {
  if (n == 0) {
    return 1;
  }
  long long int p = power(a, n / 2) % 1000000007;
  p = (p * p) % 1000000007;
  if (n % 2 == 1) {
    p = (p * a) % 1000000007;
  }
  return p % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
const int N = 1e6 + 5;
vector<long long int> sieve(N, 0);
void si() {
  sieve[1] = 1;
  for (int i = 2; i < N; i++) {
    if (sieve[i] == 0) {
      sieve[i] = 1;
      for (int j = 2 * i; j < N; j += i) {
        sieve[j] = i;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<pair<long long int, long long int> > a(n);
  vector<pair<long long int, long long int> > b, c;
  map<pair<long long int, long long int>, long long int> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    if (a[i].first < a[i].second) {
      b.push_back(make_pair(a[i].second, a[i].first));
    } else {
      c.push_back(make_pair(a[i].second, a[i].first));
    }
    m[make_pair(a[i].first, a[i].second)] = i + 1;
  }
  if ((long long int)(b.size()) > 0) {
    sort(b.begin(), b.end(), greater<pair<long long int, long long int> >());
  }
  if ((long long int)(c.size()) > 0) {
    sort(c.begin(), c.end());
  }
  for (int i = 0; i < (long long int)(b.size()); i++) {
    swap(b[i].first, b[i].second);
  }
  for (int i = 0; i < (long long int)(c.size()); i++) {
    swap(c[i].first, c[i].second);
  }
  vector<long long int> ans1, ans2;
  long long int maxi = 0;
  for (int i = 0; i < (long long int)(b.size()); i++) {
    if (i == 0) {
      ans1.push_back(m[make_pair(b[i].first, b[i].second)]);
      maxi = b[i].second;
    } else {
      if (b[i].first < maxi) {
        ans1.push_back(m[make_pair(b[i].first, b[i].second)]);
        maxi = b[i].second;
      }
    }
  }
  long long int mini = -1;
  for (int i = 0; i < (long long int)(c.size()); i++) {
    if (i == 0) {
      ans2.push_back(m[make_pair(c[i].first, c[i].second)]);
      mini = c[i].second;
    } else {
      if (c[i].first > mini) {
        ans2.push_back(m[make_pair(c[i].first, c[i].second)]);
        mini = c[i].second;
      }
    }
  }
  if ((long long int)(ans1.size()) > (long long int)(ans2.size())) {
    cout << (long long int)(ans1.size()) << endl;
    ;
    for (long long int i = 0; i < (long long int)(ans1.size()); i++)
      cout << ans1[i] << " ";
    ;
    cout << endl;
  } else {
    cout << (long long int)(ans2.size()) << endl;
    ;
    for (long long int i = 0; i < (long long int)(ans2.size()); i++)
      cout << ans2[i] << " ";
    ;
    cout << endl;
  }
  return 0;
}