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 double PI = 3.1415926535897923846;
const long long int MOD = 1000000007;
const long long int N = 998244353;
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return res;
}
long long int modinverse(long long int a) { return power(a, MOD - 2); }
bool sortcol(const pair<long long int, long long int> &a,
             const pair<long long int, long long int> &b) {
  return a.second > b.second;
}
void solve() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int> > seq1, seq2;
  map<pair<long long int, long long int>, long long int> mp1;
  map<long long int, pair<long long int, long long int> > mp2;
  vector<pair<long long int, long long int> > num(n);
  for (long long int i = 0; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    num[i] = {x, y};
    if (x > y) seq2.push_back({x, y});
    if (y > x) seq1.push_back({x, y});
    mp1[{x, y}] = i + 1;
    mp2[i + 1] = {x, y};
  }
  sort(seq1.begin(), seq1.end(), sortcol), sort(seq2.begin(), seq2.end());
  vector<long long int> ans, cur;
  for (long long int i = 0; i < (long long int)seq1.size(); i++)
    cur.push_back(mp1[seq1[i]]);
  vector<long long int> cur2;
  for (long long int i = 0; i < (long long int)seq2.size(); i++)
    cur2.push_back(mp1[seq2[i]]);
  if (cur2.size() > cur.size())
    ans = cur2;
  else
    ans = cur;
  cout << ans.size() << endl;
  for (auto x : ans) cout << x << ' ';
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  while (test--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}