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 int OO = 1e8 + 9, N = (1 << 17) + 3;
const long long mod = 998244353;
bool cmp(pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) {
  return a.first.second > b.first.second;
}
vector<int> solve(vector<pair<pair<int, int>, int> > &arr) {
  vector<int> ret;
  sort(arr.begin(), arr.end(), cmp);
  int mx = 1e6;
  for (auto &i : arr) {
    if (i.first.first < mx) ret.push_back(i.second), mx = i.first.second;
  }
  return ret;
}
vector<int> solve2(vector<pair<pair<int, int>, int> > &arr) {
  vector<int> ret;
  sort(arr.begin(), arr.end(), cmp);
  int mx = 0;
  for (auto &i : arr) {
    i.first.second *= -1;
    if (i.first.first > mx) ret.push_back(i.second), mx = i.first.second;
  }
  return ret;
}
void test() {
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int> > a, b;
  for (int i = 0; i < n; i++) {
    pair<int, int> x;
    cin >> x.first >> x.second;
    if (x.first < x.second)
      a.emplace_back(x, i);
    else
      b.push_back({{x.first, -x.second}, i});
  }
  auto v = solve(a), v2 = solve2(b);
  if (v.size() < v2.size()) swap(v, v2);
  printf("%zu\n", v.size());
  for (auto ele : v) printf("%d%c", ele + 1, " \n"[ele == v.back()]);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t = 1;
  for (int cas = 1; cas <= t; cas++) test();
}