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 "-O3"
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using str = string;
using pll = pair<ll, ll>;
inline ll divUp(ll a, ll b) { return (a + b - 1) / b; }
ll n;
vector<tuple<ll, ll, ll>> lessP, moreP;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (ll i = 0; i < n; i++) {
    ll a, b;
    cin >> a >> b;
    if (a < b)
      lessP.emplace_back(make_tuple(a, b, i));
    else
      moreP.emplace_back(make_tuple(a, b, i));
  }
  if (lessP.size() > moreP.size()) {
    sort(lessP.begin(), lessP.end(),
         [](tuple<ll, ll, ll> &a, tuple<ll, ll, ll> &b) {
           return get<1>(a) > get<1>(b);
         });
    cout << lessP.size() << "\n";
    for (auto &it : lessP) cout << get<2>(it) + 1 << "\n";
  } else {
    sort(moreP.begin(), moreP.end(),
         [](tuple<ll, ll, ll> &a, tuple<ll, ll, ll> &b) {
           return get<0>(a) < get<0>(b);
         });
    cout << moreP.size() << "\n";
    for (auto &it : moreP) cout << get<2>(it) + 1 << "\n";
  }
  return 0;
}